Eamcet Mathematics - Conic Sections Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
The equation of the parabola, whose axis is parallel to Y - axis and which passes through the points (0, 4), (1, 9) and (-2, 6), is
- (a)
\(y=2\left( x+\frac { 3 }{ 4 } \right) ^{ 2 }+\frac { 23 }{ 8 } \)
- (b)
\(y=2\left( x+\frac { 3 }{ 2 } \right) ^{ 2 }-\frac { 1 }{ 2 } \)
- (c)
\(y=2\left( x+\frac { 3 }{ 5 } \right) ^{ 2 }+\frac { 1 }{ 2 } \)
- (d)
\(y=2\left( x+\frac { 1 }{ 2 } \right) ^{ 2 }-\frac { 1 }{ 2 } \)
The equation of the parabola whose axis is parallel to Y-axis is
y=ax2+bx+c
which passes through the points (0, 4), (1, 9) and (-2, 6).
4=ax02+bx0+c ⇒ c=4
9=a+b+c ⇒ a+b=5
and 6=4a-2b+c
⇒ 4a-2b=2
⇒ 2a-b=1
3a=6
a=2
and b=3
Thus, required equation of parabola is
y=2x2+3x+4
=2\(\left(x+\frac{3}{4}\right)^{2}+\frac{23}{8}\)
If P is a point on the ellipse \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 25 } =1\) whose foci are S and S', then PS + PS' is equal to
- (a)
8
- (b)
7
- (c)
5
- (d)
10
Given, equation of ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 25 } =1\)
Here, \({ a }^{ 2 }=16,{ b }^{ 2 }=25\)
i.e., b > a
Therefore PS + PS' = 2b
[Therefore sum of focal radii = 2b]
= \(2\times 5=10\)
Observe the following columns.
ColumnI |
ColumnII |
||
A. | An ellipse passing through the origin has its foci (3, 4) and (6, 8), then length of its minor axis is | p | 8 |
B. | If PQ is focal chord of ellipse \(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 16 } =1\) which passes through S = (3 , 0) and PS = 2, then length of chord PQ is. | q | \(10\sqrt { 2 } \) |
C. | If the line y = x + K touches the ellipse 9x2 + 16y2 = 144, then the difference of values of K is | r | 10 |
D. | Sum of distances of a point on the ellipse \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1\) from the foci. | s | 12 |
- (a)
A B C D p q r s - (b)
A B C D q r r p - (c)
A B C D s r q p - (d)
None of the above
A. Points are O(0,0), P(3,4) and Q(6,8)
2a=OP+OQ=5+10=15 ⇒ a=\(\frac{15}{2}\)
Also, distance between foci,
2ae=\(\sqrt{(6-3)^{2}+(8-4)^{2}}\)=5 ⇒ e=\(\frac{1}{3}\)
⇒ b2=\(\frac{225}{4}\left(1-\frac{1}{9}\right)=50\)
⇒ b=5√2 ⇒ 2b=10√2
B. We know that
\(\frac{1}{SP}+\frac{1}{SQ}=\frac{2a}{b^{2}}\)
⇒ \(\frac{1}{2}+\frac{1}{SQ}=\frac{10}{16}\)⇒ SQ=8
⇒ PQ=10
C. If the line y=x+k touches the ellipse 9x2+16y2=144, then
k2=16(1)2+9 ⇒ k=± 5
D. Sum of the distances of a point on the ellipse from the foci=2b=8
Hence, A⇾q, B⇾r, C⇾r, D⇾p
A variable straight line of slope a intersects the hyperbola xy = 1 at two points. Find the locus of the point which divides the line segment between these points in the ratio 1 : 2.
- (a)
\({ 16x }^{ 2 }+{ y }^{ 2 }+10xy=2\)
- (b)
\({ 16x }^{ 2 }+{ 2y }^{ 2 }+10xy=2\)
- (c)
\({ 16x }^{ 2 }+{ y }^{ 2 }+9xy=2\)
- (d)
\({ 16x }^{ 2 }+{ 3y }^{ 2 }+10xy=2\)
Consider the line y=4x+c
It meets the curve xy=1at x(4x+c)=1
⇒ 4x2+cx-1=0
⇒ x1+x2=\(\frac{-c}{4}\)
Also, y(y-c)=4
⇒ y2-cy-4=0
⇒ y1+y2=c
Let the point which divides the line segment in the ratio 1:2 be (h,k)
⇒ \(\frac{x_{2}+2x_{1}}{3}\)=h ⇒ x1=3h+\(\frac{c}{4}\)
⇒ x2=-\(\frac{c}{2}\)-3h
Also, \(\frac{y_{2}+2y_{1}}{3}\)=k⇒ y1=3k-c
⇒ y2=-3k+2c
Now, (h,k) lies on the line y=4x+c
⇒ k=4h+c⇒ c=k-4h
⇒ x2=-\(\frac{k}{2}\)+2h-3h=-h-\(\frac{k}{2}\)
and y2=-3k+2k-8h=-k-8h
⇒ \(\left(h+\frac{k}{2}\right)\)(k+8h)=1
⇒ hk+8h2+\(\frac{k^{2}}{2}\)+4hk=1
⇒ 16h2+k2+10hk=2
Hence, the locus of (h,k) is 16x2+y2+10xy=2
If f(x) = ax3 of bx2 + Cx + d, (a, b, c, d are rational nos.) and roots of f(x) = 0 are eccentricities of a parabola and a rectangular hyperbola then a + b + c + d equals
- (a)
-1
- (b)
0
- (c)
1
- (d)
data inadequate
We know that eccentricity of a parabola and rectangular hyperbola are 1 and √2 respectively. Also irrational roots occur in conjugate pair, thus roots of f(x) = 0 are 1, \(\sqrt2\) and -\(\sqrt2\)
\(\therefore\) f(x) = (x -1)(x - \(\sqrt2\))(x +\(\sqrt2\))
= x3 - x2 - 2x + 2
\(\therefore\) a + b + c +d = 1 - 1 - 2 + 2 = 0
Two conics \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) and x2 = \(\frac { a }{ b } y\) intersect, if
- (a)
\(0\le\)\(1\over2\)
- (b)
0\(1\over2\)
- (c)
a22
- (d)
a2>b2
Eliminating x, we have
\(\frac { { y }^{ 2 } }{ { b }^{ 2 } } +\frac { y }{ ab } +1=0\)
This equation has real and distinct roots
\(\therefore \frac { 1 }{ { a }^{ 2 }{ b }^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } >0\)
ie,. \(\therefore\frac { 1 }{ { a }^{ 2 }{ b }^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } >0\)
\(\Rightarrow\) a < \(1\over2\) and hence the conics intersect if 0 < a < \(1\over2\)
If the sum of the slopes of the normals rrom a point P on hyperbola xy = c2 is constant k (k > 0), then the locus of P is
- (a)
y2 = k2c
- (b)
x2 = kc2
- (c)
y2 = ck2
- (d)
x2 = ck2
Equation of normal at \(\left( ct,\frac { c }{ t } \right) \) is
xt3 - yt - ct4 + c = 0
slope is t2 = k
Let x = ct and y = \(c\over t\)
ten \(x\over y\) = t2 = k (from Eq.(i))
\(\therefore \) x = ky
x2 = kxy
x2 = kc2
The line x cas a .+ y sin a = p touches the hyperbola
- (a)
a2 cos2 \(\alpha\) _.b2 sin2 \(\alpha\)= p2
- (b)
a2 cos2 \(\alpha\) - b2 sin2 \(\alpha\)= p
- (c)
a2 cos2 \(\alpha\) + b2 sin2 \(\alpha\)= p2
- (d)
a2 cos2 \(\alpha\) + b2 sin2 \(\alpha\)= p
Given line is
x cos \(\alpha\) + y sin \(\alpha\) = p
\(\Rightarrow\) y = sin\(\alpha\) = -x cos\(\alpha\) + p
\(\Rightarrow\) \(y=-\frac { cos\alpha }{ sin\alpha } x+\frac { p }{ sin\alpha } \)
Comparing with y = mx + c
\(\therefore\) m = - \(\frac { cos\alpha }{ sin\alpha } \)
and c = \(\frac { p }{ sin\alpha } \)
Condition of tangency
c2 = a2m2-b2
\(\Rightarrow\) \(\frac { { p }^{ 2 } }{ { sin }^{ 2 }\alpha } =\frac { { a }^{ 2 }{ cos }^{ 2 }\alpha }{ { sin }^{ 2 }\alpha } -{ b }^{ 2 }\)
\(\Rightarrow\) p2 = a2 cos2 \(\alpha\) - b2 sin2 \(\alpha\)
The set of postive value m for which a line with slope m is a common tangent of ellipse \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) and parabola Y2 =4ax is given by
- (a)
(2,0)
- (b)
(3,5)
- (c)
(0,1)
- (d)
None of these
Equation of tangent of y2 =4ax in terms of slope (m) is \(y=mx+\frac { a }{ 3 } \)
Which is also tangent of \(\left( \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1 \right) \)
then \(\left( \frac { a }{ 3 } \right) ^{ 2 }={ a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 }\)
\({ a }^{ 2 }\left( \frac { 1 }{ { m }^{ 2 } } -{ m }^{ 2 } \right) ={ b }^{ 2 }\)
\(\left( \frac { 1 }{ { m }^{ 2 } } -{ m }^{ 2 } \right) =\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)
\(\Rightarrow \frac { (1+3^{ 2 })+1-3^{ 2 }) }{ { m }^{ 2 } } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)
\(\Rightarrow \left( \frac { 1-{ m }^{ 2 } }{ { m }^{ 2 } } \right) =\frac { { b }^{ 2 } }{ { a }^{ 2 }((1+3^{ 2 }) } >0\)
\(\frac { 1-{ m }^{ 2 } }{ { m }^{ 2 } } >0\)
\(\frac { { m }^{ 2 }-1 }{ { m }^{ 2 } } >0\)
\(0<{ m }^{ 2 }<1\)
\(m\quad (-1,0)(0,1)\)
p(a sec \(\theta\),b tan \(\theta\)) and Q (a sec \(\theta\), b tan \(\theta\)) are the ends of a focal chord of \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) ,then tan \(\theta\)/2 tan \(\theta\) / 2 equals to
- (a)
\(\frac { e-1 }{ e+1 } \)
- (b)
\(\frac { 1-e }{ 1+e } \)
- (c)
\(\frac { 1+e }{ 1-e } \)
- (d)
\(\frac { e+1 }{ e-1 } \)
Equation PQ is \(x\over a\) cos \(\left( \frac { \theta -\phi }{ 2 } \right) -\frac { y }{ b } sin\left( \frac { \theta -\phi }{ 2 } \right) \)
\(=cos\left( \frac { \theta -\phi }{ 2 } \right) \)
its passes throug (ae, 0)
\(\therefore\) e cos \(\left( \frac { \theta -\phi }{ 2 } \right) \)-0 = cos\(\left( \frac { \theta +\phi }{ 2 } \right) \)
\(\Rightarrow \frac { cos\left( \frac { \theta -\phi }{ 2 } \right) }{ cos\left( \frac { \theta +\phi }{ 2 } \right) } \frac { 1 }{ e } \)
\(\frac { cos\left( \frac { \theta -\phi }{ 2 } \right) -cos\left( \frac { \theta +\phi }{ 2 } \right) }{ cos\left( \frac { \theta -\phi }{ 2 } \right) +cos\left( \frac { \theta +\phi }{ 2 } \right) } =\frac { 1-e }{ 1+e } \)
(by componendo and dividendo method)
\(\Rightarrow \quad tan\theta /2tan\theta /2=\frac { 1-e }{ 1+e } \)
the locus of the middle points of chords of hyperbola 3x2 - 2y2 + 4x - 6y = 0 parallel to y = 2x is
- (a)
3x - 4y = 4
- (b)
3y - 4x + 4 = 0
- (c)
4x - 4y = 3
- (d)
3x - 4y =.2
Let the middle point of the chord is (h, k)
\(\therefore\) T = S1
3xh = 2yk + 2(x + h) - 3(y + k) = 3h2 - 2k2 + 4h - 6k
Solpe of this chord = \(\frac { 3h\quad +\quad 2 }{ 2k\quad +\quad 3 } =2\) (given)
or 3h + 2 = 4k + 6
3h - 4k = 4
hence locus of middle point is
3x - 4y = 4
The distances from the foci of (p (a,b) on the ellipse \(\frac { { x }^{ 2 } }{ { 9 } } +\frac { { y }^{ 2 } }{ 25 } =1\) are
- (a)
\(\quad 4\pm \frac { 4 }{ 5 } b\)
- (b)
\(5\pm \frac { 4 }{ 5 } a\)
- (c)
\(5\pm \frac { 4 }{ 5 } b\)
- (d)
None of these
\(\frac { { x }^{ 2 } }{ { 9 } } +\frac { { y }^{ 2 } }{ 25 } =1\)
If e is eccentricity, then
9=25(1 - e2)
e=\(\frac { 4 }{ 5 } \)
\(\therefore\) foci (S) \(\equiv \)(0,\(\pm \)4)
then SP=5\(\pm \)e(b)
=5\(\pm \)\(\frac { 4b }{ 5 } \)
If x = 9 is the chord of contact of the hyperbola X2 - Y2 = 9, then the equation of the corresponding pair of tangents is
- (a)
9x2 - 8y2 + 18x - 9 = O
- (b)
9x2 - 8y2 - 18x + 9 = 0
- (c)
9x2 - 8y2 - l8x - 9 = 0
- (d)
9x2 - 8y2 + 18x + 9 = O
Since, chord of contact x = 9 which is parallel to y-axis. Hence, points of intersection of tangents must lie on x-axis. Let (h, 0) is the point of tangent, then equation of chord of
contact W.r.t. (h, 0) is|
h.x - y.0 = 9
hx = 9 but given x = 9
h=1
Coordinates of D is (1, 0)
Hence combined equation of DA and DB is
SS1 = T2
(x2 - y2 - 9)(1 - 0 - 9) = (1.x + 0.y -9)2
Then we get
9x2 + 8y2 - 18x + 9 = 0
The points of intersection of the curves whose parametric equations are x = t2 + 1, y = 2t and x = 2s, Y = 2/s, . IS given
- (a)
(1, -3)
- (b)
(2,2)
- (c)
(-2, 4)
- (d)
(1, 2)
\(\because\) x = t 2 + 1, y = 2t
\(\Rightarrow\) x = \({ \left( \frac { y }{ 2 } \right) }^{ 2 }\) + 1
\(\Rightarrow\) x = \(\frac { { y }^{ 2 } }{ 4 } \) + 1
\(\Rightarrow\) y2 = 4(x - 1)
Also, x = 2s, y = 2/s
\(\therefore\) xy = 4
From Eqs. (i) and (ii)
\({ \left( \frac { 4 }{ x } \right) }^{ 2 }\) = 4(x - 1)
\(\Rightarrow\) 16/x2 = 4(x - 1)
\(\Rightarrow\) x3 - x2 - 4 = 0
\(\Rightarrow\) (x - 2) (x2 + x + 2) = 0
x = 2 , x2 + x + 2 \(\neq\) 0
From Eq (ii)
y = 4/2 = 2.
Point of intersection is (2, 2)
The equations to the common tangents to the two hyperbolas \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad and\quad \frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } =1\) are
- (a)
\(y=\pm x\pm \sqrt { ({ b }^{ 2 }-{ a }^{ 2 }) } \)
- (b)
\(y=\pm x\pm \sqrt { ({ a }^{ 2 }-{ b }^{ 2 }) } \)
- (c)
\(y=\pm x\pm \sqrt { ({ a }^{ 2 }-{ b }^{ 2 }) } \)
- (d)
\(y=\pm x\pm \sqrt { ({ a }^{ 2 }+{ b }^{ 2 }) } \)
Let y = mx + c be a common tangent of hyperbolas
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =\ 1\) .........(1)
and \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } = \ 1\) .........(ii)
Condition of tangency for Eq. (i) is
c2 = a2m2 - b2 ..........(iii)
and condition of tangency for Eq. (ii) is
c2 = a2 - b2m2 ...........(iv)
From Eqs. (iii) and (iv),
a2m2 - b2 = a2 - b2m2
\(\Rightarrow\) a2(m2 - 1)+ b2(m2 -1) = 0
\(\Rightarrow\) (a2 + b2)(m2 -1) = 0
\(\because\) a2 + b2 \(\neq\) 0
\(\therefore\) m2 - 1 = 0
\(\Rightarrow\) m = \(\pm\) 1
From Eq. (iii), c2 = a2 - b2
\(\therefore\) c = \(\pm\) \(\sqrt { ({ a }^{ 2 }-{ b }^{ 2 }) } \)
Hence, equations of common tangents are
\(y=\pm x\pm \sqrt { ({ a }^{ 2 }-{ b }^{ 2 }) } \)
The asymptotes of the hyperbola xy = hx + ky are
- (a)
x = k, y = h
- (b)
x = h, y = k
- (c)
x = h = h
- (d)
x = k, y = k
Given hyperbola is xy=hx+ky
\(\therefore\) Pair of asymptotes is xy - hx - ky + \(\lambda\) = 0
Its represents a pair of tangents ie, pair of straight lines, then \(\triangle\) = 0
abc + 2fgh - af2 - bg2 - ch2 = 0
0 + 2 x - k/2 x -h/2 x 1/2 - 0 - 0 - \(\lambda\) x 1/4 = 0
hk/4 - \(\lambda\)/4 = 0
\(\lambda\) = hk
From Eq. (i),
xy - hx - ky +hk = 0
x(y - h) - k(y-h) = 0
(x-k)(y-h) = 0
\(\therefore\) Asymptotes are x = k, y = h
If P(x1, y1), Q(x2,y2), R (x3,y3) and S(x4, y4) are 4 concyclic points on the rectangular hyperbola xy = c2 the coordinates of the orthocentre of the \(\triangle\) PQR are
- (a)
(x4 , - y4)
- (b)
(x4 , y4)
- (c)
(-x4 - y4)
- (d)
(-x4 , y4)
Let P, Q, R, S lie on the circle x2 + y2 + 2gx+ 2fy + c = 0 and also lies on xy = c2
Solving Eqs. (i) and (ii), then
\({ x }^{ 2 }+{ \left( \frac { { c }^{ 2 } }{ x } \right) }^{ 2 }+2gx+\frac { { 2fc }^{ 2 } }{ x } +c=0\)
\(\Rightarrow\) x4 + 2gx3 + cx2 + 2fc2x + c4 = 0
\(\therefore\) x1x2x3x4 = c4 and P \(\equiv \) (x1,y1)\(\equiv \) \(\left( { x }_{ 1 },\frac { { c }^{ 2 } }{ { x }_{ 1 } } \right) \)
Q = \(\left( { x }_{ 2 },\frac { { c }^{ 2 } }{ { x }_{ 2 } } \right) \) and R=\(\left( { x }_{ 3 },\frac { { c }^{ 2 } }{ { x }_{ 3 } } \right) \)
Let orthocentre 0 \(\equiv \) (h, k)
Then, slope of QR x slope of OP = -1
\(\left( \frac { \frac { { c }^{ 2 } }{ { x }_{ 3 } } -\frac { { c }^{ 2 } }{ { x }_{ 2 } } }{ { x }_{ 3 }-{ x }_{ 2 } } \right) \times \left( \frac { k-\frac { { c }^{ 2 } }{ { x }_{ 1 } } }{ h-{ x }_{ 1 } } \right) =-1\)
\(\Rightarrow -\frac { { c }^{ 2 } }{ { { x }_{ 2 }x }_{ 3 } } \times \left( \frac { k-\frac { { c }^{ 2 } }{ { x }_{ 1 } } }{ h-{ x }_{ 1 } } \right) =-1\)
\( \Rightarrow k-\frac { { c }^{ 2 } }{ { x }_{ 1 } } =\frac { { { hx }_{ 2 }x }_{ 3 } }{ { c }^{ 2 } } -\frac { { { x }_{ 1 }{ x }_{ 2 }x }_{ 3 } }{ { c }^{ 2 } } \)
Also, slope of PQ x slope of OR = - 1
\(k-\frac { { c }^{ 2 } }{ { x }_{ 3 } } =\frac { { { hx }_{ 1 }x }_{ 2 } }{ { c }^{ 2 } } -\frac { { { x }_{ 1 }{ x }_{ 2 }x }_{ 3 } }{ { c }^{ 2 } } \)
From Eqs. (iii) and (iv),
\(\therefore\) \(h=-\frac { { c }^{ 4 } }{ { { x }_{ 1 }{ x }_{ 2 }x }_{ 3 } } \quad and\quad k=-\frac { { { x }_{ 1 }{ x }_{ 2 }x }_{ 3 } }{ { c }^{ 2 } } \)
From Eq. (iii),
h = -x4 and k = -\(\frac { { c }^{ 2 } }{ { x }_{ 4 } } \)
\(\therefore\) Orthocentre lies on xy = c2
ie, (x4,y4) and (-x4, -y4)
The domain of value of r (r > 0) for Which the above Question is true is
- (a)
r \(\epsilon \) (1, 2)
- (b)
r \(\epsilon \) (3, 4)
- (c)
r \(\epsilon \) (4, 5)
- (d)
r \(\epsilon \) (5, 6)
\(\frac { { r }^{ 2 }-9 }{ 16-{ r }^{ 2 } } >0\)
\(9<{ r }^{ 2 }<16\)
\(r\epsilon (3,4)\)
The eccentricity of the ellipse which meets the straight line\(\frac { x }{ 7 } +\frac { y }{ 2 } =1\) on the axis of x and the straight line on the \(\frac { x }{ 3 } -\frac { y }{ 5 } =1\) axis of y and whose axes lie along the axes of coordinates , is
- (a)
\(\frac { \sqrt [ 3 ]{ 2 } }{ 7 } \)
- (b)
\(\frac { \sqrt [ 2 ]{ 3 } }{ 7 } \)
- (c)
\(\frac { \sqrt { 3 } }{ 7 } \)
- (d)
None of these
Let ellipse be \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\therefore \) Ellipse meets the straight line \(\frac { x }{ 7 } +\frac { y }{ 2 } =1\) on the axis x Then \(y=0,\quad \frac { x }{ 7 } =1\)
\(\Longrightarrow \) x = 7
a = 7
and meets the straights line \(\frac { x }{ 3 } -\frac { y }{ 5 } =1\) on the axis of y, then \(x=0,0-\frac { y }{ 5 } =1\)
\(\Longrightarrow \quad y=-5\)
\(-b=-5\)
\(\therefore \quad \quad ={ b }^{ 2 }={ a }^{ 2 }(1-{ e }^{ 2 })\)
\(\Longrightarrow 25=49(1-{ e }^{ 2 })\)
\(\Longrightarrow 49{ e }^{ 2 }=24\)
\({ e }^{ 2 }=\frac { 24 }{ 49 } \)
\(\therefore \quad e=\frac { \sqrt [ 2 ]{ 6 } }{ 7 } \)
If the normals at (xi, yi), i = 1, 2, 3, 4 on the rectangular hyperbola .xy = c2, meet atthe point (\(\alpha, \beta\)). The value of \(\sum { { y }_{ i } } \)
- (a)
c\(\beta\)
- (b)
c\(\alpha\)
- (c)
\(\alpha\)
- (d)
\(\beta\)
Let (xi, yi) = \(\left( ct,\frac { c }{ t } \right) \) i = 1, 2, 3, 4 are the points on the rectangular hyperbola xy = c2
Equation of normal to the hyperbola
xy = c2 at \(\left( ct,\frac { c }{ t } \right) \) is
ct4 - t3x + ty - c = 0
It passes through (\(\alpha, \beta\)).Then
ct4 - t3\(\alpha\) +t\(\beta\) - c = 0
its biquadratic equation in t. let roots of this equations are t1, t2, t3, t4 then
\(\sum { { t }_{ 1 } } =\frac { \alpha }{ c } \) ................(1)
\(\sum { { t }_{ 1 }{ t }_{ 2 } } =0\) .............(ii)
\(\sum { { t }_{ 1 }{ t }_{ 2 }{ t }_{ 3 } } =-\beta /cand{ t }_{ 1 }{ t }_{ 2 }{ t }_{ 3 }{ t }_{ 1 }{ t }_{ 2 }{ t }_{ 4 }=-1\) ...........(iii)
\(\sum { { y }_{ i } } =\sum { \frac { 1 }{ { t }_{ 1 } } } =c\left( \frac { \sum { { t }_{ 1 }{ t }_{ 2 }{ t }_{ 3 } } }{ { t }_{ 1 }{ t }_{ 2 }{ t }_{ 3 }{ t }_{ 4 } } \right) \)
\( \\ =c\left( \frac { -\beta }{ c(-1) } \right) =\beta\)
A tangent to the ellipse 4x2+9y2 = 36 is cut by the tangent the extermities of the major axis at T and T the circle on TT' as dianeter passes through the point
- (a)
\((-\sqrt { 5 } ,0)\)
- (b)
\((\sqrt { 5 } ,0)\)
- (c)
(0,0)
- (d)
(3,2)
Any point on the ellipse Is P (3cos\(\theta\), 2sin\(\theta\)), Equation of the tangent at P is
\(\frac { x }{ 3 } cos\theta +\frac { y }{ 2 } sin\theta =1\)
Which means the Tangents x = 3 and x = -3 at the extremities of the major axis is
\(T\left( 3,\frac { 2(1-cos\theta }{ sin\theta } \right) andT'\left( 3,\frac { 2(1+cos\theta }{ sin\theta } \right) \)
Equation of the circle On TT' as diameter is
\((x-3)(x+3)+\left( y-\frac { 2(1-cos\theta }{ sin\theta } \right) \left( y-\frac { 2(1+cos\theta }{ sin\theta } \right) \)
\(\Rightarrow \quad { x }^{ 2 }+{ y }^{ 2 }-\frac { 4 }{ sin\theta } y-5=0\), Which passes through \((\pm \sqrt { 5 } ,0)\)
Two perpendicular tangents PA and PB are drawn to y2 = 4ax, minimum length of AB is equal to
- (a)
a
- (b)
4a
- (c)
8a
- (d)
2a
Chord of contact of mutually perpendicular tangents is always a focal chord. Therefore the minimum length of AB is 4a.
The standard equation of an ellipse and the general equation of a circle are respectively given by the equations.\(E:\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -1=0\) and \(c:{ x }^{ 2 }+{ y }^{ 2 }2gx+2fy+c=0\)
then the Equation \(E\quad +\lambda C=o,\lambda \neq 0\) \(\left( \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -1 \right) +\lambda ({ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c)=0\)represents a curve which passes through the common points of the ellipse and the circle The radius of the circle passing through the points of intersection E and bisectors of the Quadrants is
- (a)
\(\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \)
- (b)
\(\frac { ab\sqrt { 2 } }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \)
- (c)
\(\frac { 2ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \)
- (d)
\(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \)
\(\therefore\) Bisectors of Quadrants are y = x
and ellipse E is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) Solving Eqs. (i) and (II)
\(x=\pm \frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } y=\pm \frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \)
\(p=\left( \frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } ,\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \right) \quad Q=\left( -\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } ,\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \right) \)
\(r=\left( -\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } ,\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \right) \)and
\(S=\left( -\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } -\frac { ab }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \right) \)
\(\therefore\) P,Q,R ,S lie on a circle x2 +y2 = r2
Where r = OP = OQ = OR = OS
=\(\frac { ab\sqrt { 2 } }{ \sqrt { { (a }^{ 2 }+{ b }^{ 2 }) } } \), Where O is center of Ellipse E
The equation of the parabola whose vertex and focus lie on the axis of x, at distances a and a1 from the origin respectively is
- (a)
y2=4(a1-a)x
- (b)
y2 = 4(a1-a) (x-a)
- (c)
y2 = 4(a1 - a) (x - a1)
- (d)
none of these
Vertex is (a, 0) and focus is (a1, 0)
\(\therefore\) Distance between vertex and focus is (a1 - a)
\(\therefore\) Length of latus-rectum = 4 (a1 - a)
Equation of parabola with vertex (0, 0). latus rectum 4A and axis parallel to x -axis is
y2 = 4Ax
Now, equation of parabola with vertex (a, 0) and latus-rectum 4 (a1 - a) and axis along x-axis is
y2 = 4 (a1 - a)(x - a)
If tangents at A and B on the parabola y2 = 4ax interset at point C, then ordinates of A, C and B are
- (a)
always in AP
- (b)
always in GP
- (c)
always in HP
- (d)
none of these
Let A \(\equiv\) (at12 , 2at1), B\(\equiv\)(at22 , 2at2)
then, C \(\equiv\) (at1t2, a (t1 + t2))
\(\because\) ordinates of A, B, C are 2at1, 2at2 and a(t1+t2)
also \(\frac { ordinate\quad of\quad A+ordinate\quad of\quad B }{ 2 } =ordinate\quad 0f\quad C\)
Hence, ordinates of A, C and B are in AP.
P is a point which moves in the x-y plane such that the point P is nearer to the centre' of a square than any of the sides. The four vertices of the square are (±a,± a). The region in which p-will move is bounded by parts of parabolas of which one has the equation
- (a)
y2=a2+2ax
- (b)
x2=a2+2ay
- (c)
y2+2ax=a2
- (d)
none of these
If P≡(x,y), then
\(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)<|a-x|, \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)<|a+x|
\(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)<|a-y| and \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)<|a+y|
On squaring then we get
∴ The region bounded by the curves
x2+y2=(a-x)2,x2+y2=(a+x)2
x2+y2=(a-y)2,x2+y2=(a+y)2
The normals to the parabola y2=4ax from the point (5a,2a) are
- (a)
y=x-3a
- (b)
y=-2x+12a
- (c)
y=-3x+33a
- (d)
y=x+3a
Any normal to parabola y2=4ax may be taken as
y-2at=-t(x-at2)
or y+tx-2at-at3=0 .......(i)
If Eq(i) passes through (5a,2a), then
2a+5at-2at-at3=0
or t3-3t-2=0
or (t+1)(t2-t-2)=0
⇒ t=-1,2,-1
∴ From Eq (i) normals are
y-x+3a=0 and y+2x-12a=0
The set of points on the axis of the parabola y2 = 4x + 8 from which the 3 normals to the parabola are all real and distinct is
- (a)
\(\left\{ \left( k,0 \right) :k\le -2 \right\} \)
- (b)
\(\left\{ \left( k,0 \right) :k>-2 \right\} \)
- (c)
\(\left\{ \left( 0,k \right) :k>-2 \right\} \)
- (d)
none of these
y2 = 4 (x + 2)
let x + 2 = X and y = Y, then
Y2 = 4X ...(i)
Comparing, Eq. (i) with Y2 = 4aX
\(\therefore \) a=1
\(\because\) Three normals to the parabola are real and distinct
\(\therefore\) X >2a
\(\Rightarrow \) x+2>2
\(\Rightarrow \) x > 0
and for axis of parabola Y = 0
\(\Rightarrow \) y = 0
The set of points on the axis of parabola is (k, 0); k > 0
The length of the latus rectum of the parabola 169 {(x- 1)2 + (y - 3)2} = (5x - 12y + 17)2 is
- (a)
14/13
- (b)
12/13
- (c)
28/13
- (d)
none of these
169 [(x - 1)2 + (y - 3)2] = \({ 13 }^{ 2 }{ \left( \frac { 5x-12y+17 }{ 13 } \right) }^{ 2 }\)
\(\Rightarrow \quad { \left( x-1 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }={ \left( \frac { 5x-12y+17 }{ 13 } \right) }^{ 2 }\)
SP2 =PM2 \(\Rightarrow\) SP=PM
\(\therefore\) Focus (1, 3) directrix 5x -12y + 17 = 0
The distance of the focus from the directrix
\(=\left| \frac { 5-36+17 }{ \sqrt { 25+144 } } \right| =\frac { 14 }{ 13 } \\ \therefore \quad latusrectum=2\times \frac { 14 }{ 13 } =\frac { 28 }{ 13 } \)
Let the line Ix + my = 1 cut the parabola y2 = 4ax in the points A and B. Normals at A and B meet at point C. Normal from C other than these two meet the parabola at D, then the coordinate of D is
- (a)
(a,2a)
- (b)
\(\left( \frac { 4am }{ { l }^{ 2 } } ,\frac { 4a }{ l } \right)\)
- (c)
\(\left( \frac { 2a{ m }^{ 2 } }{ { l }^{ 2 } } ,\frac { 2a }{ l } \right)\)
- (d)
\(\left( \frac { 4a{ m }^{ 2 } }{ { l }^{ 2 } } ,\frac { 4am }{ l } \right)\)
Let A \(\equiv\) (am12 , -2am1) and B \(\equiv\) (am22, - 2am2)
Now, A and B lie on Ix + my = 1
\(\Rightarrow\) I (am12) + m (-2am1) = 1 ...(i)
and I (am22) + m (-2am2) = 1 ...(ii)
Subtracting Eq. (ii) from Eq. (i), then
la (m12 - m22) - 2am (m1 - m2) = 0
\(\Rightarrow\) a(m1 - m2) = 0
l (m1 + m2) - 2m = 0
\(\Rightarrow\) m1 + m2 = \(\frac { 2m }{ l }\) ...(iii)
Let D\(\equiv\) (am32 - 2am3) and C\(\equiv\) (h,k)
\(\therefore\) Equation of normal in terms of slope
y = Mx - 2aM - aM3
then aM3 - (h - 2a) M + k = 0
\(\therefore\) m1 + m2 + m3 = 0 \(\Rightarrow\) \(\frac { 2m }{ l }\)+ m3 = 0
\(\therefore \quad { m }_{ 3 }=-\frac { 2m }{ l } \\ \therefore \quad \quad D\equiv \left( a{ \left( \frac { -2m }{ l } \right) }^{ 2 },-2a\left( \frac { -2m }{ l } \right) \right) \\ \Rightarrow \quad D\equiv \left( \frac { 4a{ m }^{ 2 } }{ { l }^{ 2 } } ,\frac { 4am }{ l } \right) \)
If the normal at P 't' on .y2 = 4ax meets the curve again at Q, the point on the curve, the normal at which also passes through Q has coordinates
- (a)
\(\left( \frac { 2a }{ { t }^{ 2 } } ,\frac { 2a }{ t } \right)\)
- (b)
\(\left( \frac { 4a }{ { t }^{ 2 } } ,\frac { 2a }{ t } \right)\)
- (c)
\(\left( \frac { 4a }{ { t }^{ 2 } } ,\frac { 4a }{ t } \right)\)
- (d)
\(\left( \frac { 4a }{ { t }^{ 2 } } ,\frac { 8a }{ t } \right)\)
\(\because\) P \(\equiv\) (at2, 2at)
and let R = (at12 ,2at1)
\(\because\) Normals at P and R meet parabola y2 = 4ax at Q, then
tt1 = 2
or t1 = \(\frac { 2 }{ t }\)
\(\therefore \quad R\equiv \left( a{ \left( \frac { 2 }{ t } \right) }^{ 2 },2a\left( \frac { 2 }{ t } \right) \right) \\ \therefore \quad R\equiv \left( \frac { 4a }{ { t }^{ 2 } } ,\frac { 4a }{ t } \right) \)
The locus of point of intersection of tangents to the parabolas y2 = 4 (x + 1) and y2 = 8 (x + 2) which are perpendicular to each other is
- (a)
x + 7 = 0
- (b)
x - y = 4
- (c)
x + 3 = 0
- (d)
y - x = 12
\(y=m(x+1)+\left( { 1 }/{ m } \right) \\ or\quad y=mx+\left( m+\frac { 1 }{ m } \right) \quad \quad \quad ...(i)\\ is\quad a\quad tangent\quad to\quad the\quad first\quad parabola\\ and\quad y={ m }^{ \prime }\left( x+2 \right) +\frac { 2 }{ { m }^{ \prime } } \\ ={ m }^{ \prime }x+2\left( { m }^{ \prime }+\frac { 1 }{ { m }^{ \prime } } \right) \quad \quad ...(ii)\\ is\quad a\quad tangent\quad to\quad the\quad second\quad parabola\quad given\\ m.{ m\prime }=-1\\ or\quad { m\prime }=-\frac { 1 }{ m } \\ Then,from\quad Eq.(ii)\\ y=-\frac { x }{ m } +2\left( -\frac { 1 }{ m } -m \right) \\ \Rightarrow y=-\frac { x }{ m } -2\left( m+\frac { 1 }{ m } \right) \quad \quad ...(iii)\\ Subtracting\quad Eq.(iii)from\quad Eq.(i),\quad then\\ x\left( m+\frac { 1 }{ m } \right) +3\left( m+\frac { 1 }{ m } \right) =0\\ or\quad x+3=0\)
A circle passes through the points (2, -2), (3, 4) and has its centre on the line 2x+2y=7. Find its centre and radius.
- (a)
\(\left(1,{5\over2}\right),{\sqrt{37}\over 2}\)
- (b)
\((5,1), \sqrt{37}\)
- (c)
\((1,5), \sqrt{37}\)
- (d)
\(\left({5\over2},1\right),{\sqrt{37}\over 2}\)
Let the equation of the circle be
x2+y2+2gx+2fy+c=0 ....(i)
As (2, -2), (3,4) lie on the circle, we have
4+4+4g-4f+c=0⇒4g-4f+c+8=0 ...(ii)
and 9+16+6g+8f+c=0⇒6g+8f+c+25=0 ....(iii)
Since the centre (-g, -f) lies on
2x+2y-7=0, therefore, we get
-2g-2f-7=0⇒2g+2f+7=0 ..(iv)
Subtracting (ii) from (iii), we get
2g+12f+17=0 ...(v)
Subtracting (iv) from (v) we get10f+10=0⇒f=-1
putting value of f in (iv), we get 2g+5=0⇒g=-\(5\over 2\) and then (ii) gives c=-2
Also, we noe that g2+f2-c=\(25\over4\)+1+2=\(37\over 4\) which is positive
Its centre is \(\left({5\over 2},1\right)\) and radius \(=\sqrt{{25\over4}+1+2}=\sqrt{37\over 2}\)
Find the equation of the parabola with vertex at (0,0) and focus x=-2
- (a)
x2=8y
- (b)
y2=4y
- (c)
x2=4x
- (d)
y2=8x
Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis of the parabola therefore equation of the parabola is of the form x2 =4ay, thus we have x2=4(2)y, i.e. x2=8y
If the equation of the parabola is x2 =-8y, the equation of the directrix and length of latus rectum respectively are
- (a)
y=3, 4
- (b)
y=-2,4
- (c)
y=2,8
- (d)
y=-2,8
Comparing the given equation with standard form x2=-4ay, we get a=2
Therefore, the equation of directrix is y=2 and the length of the latus rectum is 4a, i.e., 8
find the coordinates of the foci and eccentricity respectively of the ellipse \({x^2\over 25}+{y^2\over 9}=1\)
- (a)
\((0, ±4),{4\over 5}\)
- (b)
\((±4,0),{4\over 5}\)
- (c)
\((0, ±4),{4\over3}\)
- (d)
\((0, ±2),{4\over5}\)
From the given equation of ellipse, we get a=5 and b=3
\(∴\ e=\sqrt{1-{b^2\over a^2}}={4\over5}\)
Therefore, the coordinates of the foci are (±ae, 0)i.e., (±4,0)
If the foci of the ellipse \({x^2\over 9}+{y^2\over 16}=1\) are (0, √7) and (0, -√7) then the foci of the ellipse \({x^2\over 9+t^2}+{y^2\over 16+t^2}=1, t\in R \), are
- (a)
(0, √7) (0, -√7)
- (b)
(0, √7) (0, -7)
- (c)
(0, 2√7) (0, -2√7)
- (d)
(√7, 0) (-√7,0)
For the equation of ellipse \({x^2\over 9}+{y^2\over 16}=1\)
Here b>a. So, its foci will be (0, ±be)
Now \(=\sqrt{1-{a^2\over b^2}}=\sqrt{1-{9+t^2\over 16+t^2}}={\sqrt7\over \sqrt{16+t^2}}\)
∴ Foci ≡ (0, √7) (0, -√7)
The equation of the ellipse centred at (1, 2) with focus (6, 2) and passing through the point (4, 6) is
- (a)
\({(x-1)^2\over 45}+{(y-2)^2\over 20}=1\)
- (b)
\({(x+1)^2\over 45}+{(y+2)^2\over 20}=1\)
- (c)
\({(x-1)^2\over 20}+{(y-2)^2\over 45}=1\)
- (d)
\({(x+1)^2\over 20}+{(y+2)^2\over 45}=1\)
The sum of the distance of a point (2, -3) from the foci of an ellipse 16(x-2)2+25(y+3)2=400 is
- (a)
8
- (b)
6
- (c)
50
- (d)
32
Given that \({(x-2)^2\over 25}+{(y+3)^2\over 16}=1\)
⇒ a2=25 ⇒ a=5 and b2=16 ⇒ b=4
Now, \(e=\sqrt{1-{b^2\over a^2}}=\sqrt{1-{16\over 25}}={3\over 5}\)
Given point (2, -3) is the centre of ellipse so sum of distance of two foci from centre is
\(2ae=2\times5\times{3\over6}=6\)
The equation \({x^2\over 1-r}-{y^2\over 1+r}=1, |r|<1\) represents a/an
- (a)
ellipse
- (b)
hyperbola
- (c)
circle
- (d)
None of these
Since |r|<1, 1-r and 1+r are both positive. So we put 1-r=a2, 1+r=b2. Then the given equation becomes, \({x^2\over a^2}-{y^2\over b^2}=1\)
The point (at2, 2bt) lies on the hyperbola \({x^2\over a^2}-{y^2\over b^2}=1\) for
- (a)
all values of t
- (b)
t2=2+√5
- (c)
t2=2-√5
- (d)
no real values of t
The point (at2, 2bt) lies on the hyperbola \({x^2\over a^2}-{y^2\over b^2}=1\)
\(∴\ {(at^2)^2\over a^2}-{(2bt)^2\over b^2}=1⇒ t^2-4t^2-1=0\)
⇒ (t2-2)2=5 ⇒ t2-2= ±√5 ⇒ t2=2±√5
But t2=2-√5 does not give real value for t, so
t2=2+√5
The eccentricity of the hyperbola \({x^2\over a^2}-{y^2\over b^2}=1\) which passes through the points (3,0) and (3,√2) is
- (a)
\(1\over \sqrt{13}\)
- (b)
\(\sqrt{13}\)
- (c)
\( \sqrt{13}\over 2\)
- (d)
\( \sqrt{13}\over 3\)
Given that the hyperbola \({x^2\over a^2}-{y^2\over b^2}=1\) passes through the points (3,0) and (3√2, 2), so we get a2=9 and b2=4
Again, we know that b2=a2(e2-1). This gives
4=9(e2-1) ⇒ e2=\({13\over 9}⇒e={\sqrt{13}\over 3}\)
The length of the transverse axis along x-axis with centre at origin of a hyperbola is 7 and it passes through the point (5, -2). The equation of the hyperbola is
- (a)
\({4\over 49}x^2-{196\over 51}y^2=1\)
- (b)
\({49\over 4}x^2-{51\over 196}y^2=1\)
- (c)
\({4\over 49}x^2-{51\over 196}y^2=1\)
- (d)
none of these
The equation of hyperbola referred to its axes as axes of co-ordinate whose distance between the foci is 20 and eccentricity equals √2 is
- (a)
x2-y2=25
- (b)
x2-y2=50
- (c)
x2-y2=125
- (d)
x2+y2=25
Given 2ae=20 and e=√2 \(∴\ a={20\over 2e}=5\sqrt2\)
Again b2=a2(e2-1) ⇒b2=25 x 2(2-1)=50
∴ Equation of hyperbola x2-y2=50
The equation of the hyperbola in the standard form (with transverse axis along the x-axis) having the length of the latus rectum = 9 units and eccentricity = 5/4 is
- (a)
\(\frac{x^2}{16}-\frac{y^2}{18}=1\)
- (b)
\(\frac{x^2}{36}-\frac{y^2}{27}=1\)
- (c)
\(\frac{x^2}{64}-\frac{y^2}{36}=1\)
- (d)
\(\frac{x^2}{36}-\frac{y^2}{64}=1\)
Let equation of hyperbola be \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(ஃ e=\sqrt{1+\frac{b^2}{a^2}}\Rightarrow (\frac{5}{4})^2=(1+\frac{b^2}{a^2})\)
\(\Rightarrow \frac{b^2}{a^2}=\frac{9}{16}\Rightarrow \frac{b}{a}=\frac{3}{4}\)
We have, \(\frac{2b^2}{a}=9 \Rightarrow 2\times \frac{3}{4}b=9\)
⇒ b = 6 and a = 8
ஃ Equation of hyperbola is \(\frac{x^2}{64}-\frac{y^2}{36}=1\)
Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
- (a)
x2 + y2 + 13y = 0
- (b)
3x2 + 3y2 + 13x + 3 = 0
- (c)
6x2+6y2-26y=0
- (d)
x2+y2+13x+3=0
Let the centre be (0, k).
The equation of circle be
x2+y2-2(0)x-2ky+c=0
⇒ x2+y2-2ky+c=0...(i)
Since, it passes through (0, 0) and (2, 3).
ஃ 02+02-2k(0)+c=0 ⇒ c=0
and (2)2 + (3)2 - 2k(3) + c = 0
⇒ 13 - 6k + c = 0 ⇒ 13 - 6k + 0 = 0 ⇒ k=\(13\over6\)
On putting the values of k and c in (i), we get
x2+y2-2\((\frac{13}{6})\)y+0=0 ⇒ 6x2+6y2-26y=0
If the centre of ellipse is at the origin, then
Statement-I: Ellipse is not symmetric with respect to X-axis or Y-axis.
Statement-II: Ellipse is symmetric with respect to both the coordinate axes
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
Ellipse is symmetric with respect to both the coordinate axes, since if (x, y) is a point on the ellipse, then (-x, y), (x, -y) and (-x, -y) are also points on the ellipse.
The equations of the lines joining the vertex of parabola y2 = 6x to the points on it which have abscissa 24, are
- (a)
y±2x=0
- (b)
2y±x=0
- (c)
x±2y=0
- (d)
2x±y=0
Let P and Q be points on the parabola y2=6x and OP, OQ be the lines joining the vertex O to the points P and Q whose abscissa is 24.
Thus y2=6 x 24 = 144 or y = ±12
Therefore the co-ordinates of the points P and Q are (24, 12) and (24, -12) respectively. Hence, the lines passing through origin are y=士\(12\over24\)x ⇒ 2y = ± x.
The locus of the intersection point of \(xcos\alpha -ysin\alpha =\alpha \) and \(xsin\alpha -ysin\alpha =b\) is (Mark 1 for ellipse 2 for hyperbola, 3 for parabola and 4 for None of these).
- (a)
3
- (b)
2
- (c)
4
- (d)
5
\(xcos\alpha -ysin\alpha =\alpha \) ,\(xsin\alpha -ysin\alpha =b\)
Intersection points are
\(h=\cfrac { acos\alpha -bsin\alpha }{ cos2\alpha } ,k=\cfrac { asin\alpha -bcos\alpha }{ cos2\alpha } \)
Then the locus of point (h,k) is x4+y4-2x2y2=(a2+b2)(x2+b2)(x2+y2) which is not a locus of any given curves.
y2-2x-2y+5=0 represents a parabola whose directrix is x=3/A.Find A.
- (a)
2
- (b)
4
- (c)
5
- (d)
6
The equation can be written as (y-1)2=2(x-2)
Obviously, it is a parabola whose focus is \(\left( \cfrac { 5 }{ 2 } ,1 \right) \) and directrix is \(x=\cfrac { 3 }{ 2 } \)