IISER Mathematics - Conic Sections
Exam Duration: 45 Mins Total Questions : 30
The eccentricity of the ellipse which meets the straight line \(\frac { x }{ 7 } +\frac { y }{ 2 } =1\) on the X - axis and the straight line \(\frac { x }{ 7 } -\frac { y }{ 2 } =1\) on the Y - axis and whose axes lie along the axes of coordinates is
- (a)
\(\frac { 1 }{ \sqrt { 2 } } \)
- (b)
\(\frac { 2\sqrt { 6 } }{ 7 } \)
- (c)
\(\frac { \sqrt { 3 } }{ 7 } \)
- (d)
\(\frac { \sqrt { 5 } }{ 7 } \)
Let the equation of the ellipse be
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
which passes through the points (7, 0) and (0, -5).
So, a2 = 49 and b2 = 25
\(\therefore \quad { b }^{ 2 }={ a }^{ 2 }(1-{ e }^{ 2 })\)
\(\\ \Rightarrow 25=49(1-{ e }^{ 2 })\)
\(\\ \Rightarrow e=\frac { 2\sqrt { 6 } }{ 7 } \)
If the line ax + by + c = 0 is a normal to the hyperbola xy = 1, Then,
- (a)
\(a=0,b\neq 0\)
- (b)
\(a>0,b<0\)
- (c)
\(a<0,b<0\)
- (d)
\(a\neq 2,b\neq 1\)
Given, equation of hyperbola is x = 1
Equation of normal at, i.e., \(\left( t,\frac { 1 }{ t } \right) \) is \({ xt }^{ 3 }-yt-{ t }^{ 4 }+1=0\)
Slope is t2 = - a/b
\(\therefore \quad -\frac { a }{ b } >0\Rightarrow \frac { a }{ b } <0\)
\(\Rightarrow \quad a>0,b<0\quad or\quad a<0,b>0\)
The equation of a tangent to the parabola y2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is
- (a)
(-1, 1)
- (b)
(0, 2)
- (c)
(2, 4)
- (d)
(-2, 0)
Since, perpendicular tangents intersect on the directrix, then point must lie on the directrix x = -2.
Hence, the required point is (-2, 0).
The set of values of a for which (13x-1)2 + (13y - 2)2 = a (5x+12 y -1)2 represents an ellipse , if
- (a)
1 < a < 2
- (b)
0 < a < 1
- (c)
2 < a < 3
- (d)
None of these
We have 13x-1)2 + (13y - 2)2 = a (5x+12 y -1)2
\(\left( x-\frac { 1 }{ 13 } \right) ^{ 2 }+\left( y-\frac { 2 }{ 13 } \right) ^{ 2 }=\frac { a(5x+12y-1)2 }{ \left( \sqrt { { (5 }^{ 2 }+{ 12 }^{ 2 }) } \right) ^{ 2 } } \)
\(or\quad \sqrt { \left( x\frac { 1 }{ 13 } \right) +\left( y-\frac { 2 }{ 13 } \right) ^{ 2 } } =\sqrt { a } \frac { |5x+12y-1| }{ \sqrt { \left( { 5 }^{ 2 }+{ 12 }^{ 2 } \right) } } \)
\(\therefore \quad SP=ePM\)
\(\Longrightarrow \quad 0<\sqrt { a } <1\)
∵
If e and e' be the eccentricities of a hyperbola and its conj.ugate, then \(\frac { 1 }{ { e }^{ 2 } } +\frac { 1 }{ { e }^{ '2 } } \) is equa I to
- (a)
0
- (b)
1
- (c)
2
- (d)
none of these
e2 = 1 + \(\frac { { b }^{ 2 } }{ { a }^{ 2 } } \) and (e')2 = 1 + \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } \)
\(\therefore\) \(\frac { 1 }{ { e }^{ 2 } } +\frac { 1 }{ { e }^{ '2 } } =\frac { { a }^{ 2 } }{ { ({ a }^{ 2 }+b }^{ 2 }) } +\frac { { b }^{ 2 } }{ { ({ a }^{ 2 }+b }^{ 2 }) } \) = 1
The diameter of 16x2 - 9y2 = l44, which is conjugate to x = 2y, is
- (a)
\(y=\frac { 16 }{ 9 } x\)
- (b)
\(y=\frac { 32 }{ 9 } x\)
- (c)
\(x=\frac { 16 }{ 9 } y\)
- (d)
\(x=\frac { 32 }{ 9 } y\)
Given hyperbola is 16x2 - 9y2 = 144
or \(\frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\)
Comparing with \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
a = 3 and b = 4
y = m1x and y = m2x are the conjugate diameters then m1m2 = \(\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)
\(\because\) x = 2y, \(\therefore\) m1 = \(1\over2\))
\(\Rightarrow\) \(1\over2\) x m2 = \(19\over6\)
or m2 = \(32\over9\)
\(\therefore\) Diameter is y = \(32\over9\) x
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 \(\left( \frac { a }{ 3 } \right) ^{ 2 }={ a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 }\)
\({ a }^{ 2 }\left( \frac { 1 }{ { m }^{ 2 } } -{ m }^{ 2 } \right) ={ b }^{ 2 }\)
\(\left( \frac { 1 }{ { m }^{ 2 } } -{ m }^{ 2 } \right) =\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)
\(\Rightarrow \frac { (1+3^{ 2 })+1-3^{ 2 }) }{ { m }^{ 2 } } =\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)
\(\Rightarrow \left( \frac { 1-{ m }^{ 2 } }{ { m }^{ 2 } } \right) =\frac { { b }^{ 2 } }{ { a }^{ 2 }((1+3^{ 2 }) } >0\)
\(\frac { 1-{ m }^{ 2 } }{ { m }^{ 2 } } >0\)
\(\frac { { m }^{ 2 }-1 }{ { m }^{ 2 } } >0\)
\(0<{ m }^{ 2 }<1\)
\(m\quad (-1,0)(0,1)\)
Product of the lengths of the perpendiculars drawn frhm foci on any tangent to the hyperbola x2/a2 - y2/b2 = 1 is
- (a)
\(1\over2\)b2
- (b)
b2
- (c)
a2
- (d)
\(1\over2\)b2
Equation of tangent hyperbola \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)at '\(\phi\)' is
\(x\over a\) sec \(\phi\) - \(y\over b\) tan \(\phi\) = 1since foci (\(\pm\)ae, 0)
p1 x p2 = \(\frac { (e\quad sec\phi -1)(e\quad sec\phi +1) }{ \left( \sqrt { \frac { { sec }^{ 2 }\Phi }{ { a }^{ 2 } } +\frac { { tan }^{ 2 }\Phi }{ { b }^{ 2 } } } \right) } \)
\(=\frac { (e\quad sec\phi -1){ a }^{ 2 }{ b }^{ 2 } }{ { (a }^{ 2 }{ tan }^{ 2 }\Phi +{ b }^{ 2 }{ sec }^{ 2 }\Phi )\quad } \)
\(=\frac { { (a }^{ 2 }{ e }^{ 2 }{ sec }^{ 2 }\Phi -{ a }^{ 2 }){ b }^{ 2 } }{ { (a }^{ 2 }{ tan }^{ 2 }\Phi +{ b }^{ 2 }{ sec }^{ 2 }\Phi )\quad } \)
\(=\frac { { [(a }^{ 2 }{ +b }^{ 2 }){ sec }^{ 2 }\Phi -{ a }^{ 2 }]{ b }^{ 2 } }{ { (a }^{ 2 }{ tan }^{ 2 }\Phi +{ b }^{ 2 }{ sec }^{ 2 }\Phi )\quad } \)
\(=\frac { { (a }^{ 2 }{ tan }^{ 2 }\Phi +{ b }^{ 2 }{ sec }^{ 2 }\Phi ){ b }^{ 2 } }{ { (a }^{ 2 }{ tan }^{ 2 }\Phi +{ b }^{ 2 }{ sec }^{ 2 }\Phi )\quad } ={ b^2 }\)
The parametric representation of a point on the ellipse whose foci are (-1, 0) and (7, 0) and eccentricity is 1/2, is
- (a)
\((3+8cos\theta ,4\sqrt { 3 } sin\theta )\)
- (b)
\((8cos\theta ,4\sqrt { 3 } sin\theta )\)
- (c)
\((3+4\sqrt { 3 } cos\theta ,8sin\theta )\)
- (d)
None of these
Foci are (-1,0) and (7, 0) coordinate of the center is (3,0) distance between focii = \(\sqrt { \left( -1-1 \right) ^{ 2 }+0 } =8\)
2ae = 8
ae = 4
Here e=\(e=\frac { 1 }{ 2 } \)
\(\therefore\) a=8
\(\therefore\) b2 -a2 (1- e2) = 64 \(\left( 1-\frac { 1 }{ 4 } \right) \)
= 64-16=48
Equation of ellipse with center (3, 0) is \(\frac { \left( x-3 \right) ^{ 2 } }{ 64 } +\frac { y^{ 2 } }{ 48 } =1\)
\(\therefore\) parametric coordinated \((3+8cos\theta ,4\sqrt { 3 } sin\theta )\)
The eccentric angles of the extremities of latus-rectum to the ellipse \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) are given by
- (a)
\(tan^{ -1 }\left( \pm \frac { ae }{ b } \right) \)
- (b)
\(tan^{ -1 }\left( \pm \frac { be }{ a } \right) \)
- (c)
\(tan^{ -1 }\left( \pm \frac { b }{ ae } \right) \)
- (d)
\(tan^{ -1 }\left( \pm \frac { a }{ be } \right) \)
Let a cos \(\theta\) =\(\pm \) ae and b sin =\(\pm \) \(\frac { { b }^{ 2 } }{ a } \)
\(\therefore\) \(\frac { b }{ a } tan\phi =\pm \frac { { b }^{ 2 } }{ { a }^{ 2 }e } \)
\(\Longrightarrow tan\phi =\pm \left( \frac { b }{ ae } \right) \)
\(\therefore \quad \phi =tan^{ -1 }\quad \left( \pm \frac { b }{ ae } \right) \)
If we rotate the axes of the rectangular hyperbola x2 -y2 = a2 through an angle \(\pi/4\) in the clockwise direction, then the equation x2 - y2 = a2 reduces to xy \(=\frac { { a }^{ 2 } }{ 2 } ={ \left( \frac { a }{ \sqrt { 2 } } \right) }^{ 2 }={ c }^{ 2 }\) (say)
Since x = ct, y = \(\frac {c}{t}\) satisfies.xy = c2. \(\therefore\) ( x, y ) = \((ct,\frac{c}{t})\)\((t\neq0)\) is called a "t' point on the rectangular hyperbola.
Let \(\alpha, \gamma\) be the roots of the equation t1X2 - 4x + 1 = 0 and \(\beta, \delta\) be the roots of t2x2 - 6x + 1 = 0 and \(\alpha,\beta,\alpha,\gamma,\delta\) are in HP then the point of intersection of normals at 't1' and 't2' on .xy = c2 is
- (a)
\((\frac {327c}{264},\frac {921c}{264})\)
- (b)
\((\frac{237c}{264},\frac{291c}{264})\)
- (c)
\((\frac {723c}{264},\frac{129c}{264})\)
- (d)
none of these
\(\because\) \(\alpha +\gamma =\frac {4}{t_1},\alpha\gamma=\frac{1}{t_1}\)
and \(\beta + \delta = \frac {6}{t_2},\beta\delta=\frac{1}{t_2}\)
Also, \(\alpha,\beta,\gamma,\delta\) are in HP
\(\therefore\) \(\beta = \frac {2\alpha\gamma}{\alpha+\gamma}=\frac{2\times\frac{1}{t_1}}{\frac{4}{t_1}}=\frac{1}{2}\) .............(i)
and \(\gamma = \frac {2\beta\delta}{\beta+\delta}=\frac{2\times\frac{1}{t_1}}{\frac{6}{t_1}}=\frac{1}{3}\) ............(ii)
Now, \(\gamma\) lie on t1x2 - 4x + 1 = 0
\(\Rightarrow\) \({ t }_{ 1 }{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }-\quad 4{ \left( \frac { 1 }{ 3 } \right) }+1=0\)
\(\Rightarrow\) t1 = 8
\(\because\) Normal at 't1' is
xt13 - yt1 - ct14 + c = 0 ...........(iii)
and normal at 't2' is
xt32 - yt2 - ct42 + c = 0 ..............(iv)
Solving Eqs, (ii) and (iv) we get point of intersection is
\(\left( \frac { c\{ { t }_{ 1 }{ t }_{ 2 }({ t }_{ 1 }^{ 2 }+{ t }_{ 1 }{ t }_{ 2 }+{ t }_{ 2 }^{ 2 })-1\} }{ { t }_{ 1 }{ t }_{ 2 }({ t }_{ 1 }+{ t }_{ 2 }) } ,\frac { c\{ { t }_{ 1 }^{ 3 }+{ t }_{ 2 }^{ 3 }+({ t }_{ 1 }^{ 2 }+{ t }_{ 1 }{ t }_{ 2 }+{ t }_{ 2 }^{ 2 })\} }{ { t }_{ 1 }{ t }_{ 2 }({ t }_{ 1 }+{ t }_{ 2 }) } \right) \)
or \(\left( \frac { 2327c }{ 264 } ,\frac { 13921c }{ 264 } \right) \)
The eccentric angle of a point on the ellipse X2 + 3y2 = 6 at point on the ellipse X2 + 3Y2 = 6 at a distance 2 unit from origin is
- (a)
\(\frac { \pi }{ 4 } \)
- (b)
\(\frac { 3\pi }{ 4 } \)
- (c)
\(\frac { 5\pi }{ 4 } \)
- (d)
\(\frac { 7\pi }{ 4 } \)
\(\frac { { x }^{ 2 } }{ 6 } +\frac { { y }^{ 2 } }{ 2 } =1\)
a point of the ellipse (i) is \(\left( \sqrt { 6 } cos\phi ,\quad \sqrt { 2 } sin\phi \right) \)Since, its distance from the origin is 2
\(\therefore 6{ cos }^{ 2 }+{ 2sin }^{ 2 }\phi =({ 2 })^{ 2 }\)
\(\Longrightarrow \quad 4{ cos }^{ 2 }\phi +2=4\)
\(\Longrightarrow \quad 4{ cos }^{ 2 }\phi =2\)
\({ \Longrightarrow \quad cos }^{ 2 }1/2\)
\({ \Longrightarrow \quad 2cos }^{ 2 }\phi =1\)
\({ cos }2\phi =0\)
\(\Longrightarrow \quad 2\phi =(2n+1)\frac { \pi }{ 2 } ;n\epsilon I\)
\(or\quad \phi =(2n+1)\pi /4;n\epsilon I\)
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 Let the eccentric angles of three points p, Q and R in the Ellipse E are\(\frac { \pi }{ 2 } +a\) and \(\pi +a\) circle C through P, Q and R cuts the ellipse E again at S, then the eccentric angle of S is
- (a)
\(\frac { \pi }{ 2 } -3a\quad \)
- (b)
\(\pi -3a\)
- (c)
\(\frac { 3\pi }{ 2 } -3a\)
- (d)
\(2\pi -3a\)
Eccentric angles Of P, Q and R are \(\frac { \pi }{ 2 } +a\) and \(\pi -a\) respectively. Let be the eccentric abgle of the fourth point S.
Then \(a+\frac { \pi }{ 2 } +a+\pi +a+\beta =2\eta \pi \)
or \(\beta =2\eta \pi -\frac { 3\pi }{ 2 } -3a\)
for \(\eta =1,\beta =\frac { \pi }{ 2 } -3a\)
The equation \(\sqrt { \left\{ { \left( x-3 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 } \right\} } +\sqrt { \left\{ { \left( x+3 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 } \right\} } =6\) represents
- (a)
an ellipse
- (b)
a pair of straight lines
- (c)
a circle
- (d)
a straight line joining the point (-3,1) to the point (3,1)
Let 1= (x - 3)2 +(y- 1)2 and m = (x + 3)2 +(y - 1)2
m-l=12x
\(\Rightarrow \quad \left( \sqrt { m } +\sqrt { l } \right) \left( \sqrt { m } -\sqrt { l } \right) =12x\\ \Rightarrow \quad \left( \sqrt { m } -\sqrt { l } \right) =2x\quad \quad ...(i)\\ and\quad \left( \sqrt { m } +\sqrt { l } \right) =6\quad \quad ...(ii)\)
Adding Eqs. (i) and (ii), then
\(2.Jm = 2(x + 3)\)
Squaring both sides m =x2 + 6x + 9
\(\Rightarrow\) x2 + y2 + 6x - 2y + 10 = x2 + 6x + 9
\(\Rightarrow\) y2 - 2y + 1 = 0
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 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
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 = \(\frac { 2 }{ t }\)
\(\therefore \quad R\equiv \left( a{ \left( \frac { 2 }{ t } \right) }^{ 2 },2a\left( \frac { 2 }{ t } \right) \right) \\ \therefore \quad R\equiv \left( \frac { 4a }{ { t }^{ 2 } } ,\frac { 4a }{ t } \right) \)
If the tangent at P on y2 = 4ax meets the tangent at the vertex in Q and S is the focus of the parabola, then \(\angle\)SQP is equal to
- (a)
\({ \pi }/{ 3 }\)
- (b)
\({ \pi }/4\)
- (c)
\({ \pi }/2\)
- (d)
\({ 2\pi }/{ 3 }\)
Let P \(\equiv\)(at2, 2at),
\(\therefore\) Tangent at P is, ty = x + at2 ...(i)
tangent at vertex is x = 0 ...(ii)
Solving Eqs. (i) and (ii), we get
Q\(\equiv\)(0, at) and S\(\equiv\)(a.,0)
\(\because\) Slope of QP is \(\frac { 2at-at }{ a{ t }^{ 2 }-0 } =\frac { 1 }{ t } ={ m }_{ 1 }\quad (say)\)
and slope of QS is \(\frac { 0-at }{ a-0 } =-t={ m }_{ 2 }\quad \quad (say)\)
\(\because\) m1m2 = -1
\(\therefore\) \(\angle\)SQP=\({ \pi }/2\)
The length of latus-rectum of the parabola whose parametric equation is x = t2 + t + 1, y = t2 -t +1, where \(t\in R\) is equal to
- (a)
2
- (b)
4
- (c)
8
- (d)
none of these
Now, t2 + 1= x - t ...(i)
and t2 + 1= y + t ...(ii)
\(\therefore \quad t=\left( \frac { x-y }{ 2 } \right)\)
From Eq. (i)
\({ \left( \frac { x-y }{ 2 } \right) }^{ 2 }+1=x-\left( \frac { x-y }{ 2 } \right)\)
\( =\left( \frac { x+y }{ 2 } \right)\)
\(\Rightarrow \quad { \left( x-y \right) }^{ 2 }+4=2\left( x+y \right)\)
\(\Rightarrow \quad { \left( x-y \right) }^{ 2 }=2\left( x+y-2 \right)\)
∴ Length of latus-rectum=2
The equation of family of circles with centre at (h, k) touching the x-axis given by
- (a)
x2+y2-2hx+h2=0
- (b)
x2+y2-2hx-2ky+h2=0
- (c)
x2+y2-2hx-2ky-h2=0
- (d)
x2+y2-2hx-h2=0
When Circle with centre (h, k) touches x-axis, the radius (r)=k
(x-h)2+(y-k)2=k2
⇒ x2+y2-2hx-2ky+h2+k2=k2
⇒ x2+y2-2hx-2ky+h2=0
The focus of the parabola y2-x-2y+2=0 is
- (a)
\(\left({1\over4},0\right)\)
- (b)
(1, 2)
- (c)
\(\left({3\over4},1\right)\)
- (d)
\(\left({5\over4},1\right)\)
We have, y2-2y=x-2 or (y-1)2=x-1. Shifting the origin at (1, 1), we have x=X+1, y=Y+1.The equation (y-1)2=(x-1) reduce toY2=X. This represents a parabola with latus rectum =1. Coordinates of the focus w.r.t new axes are \(\left( {1\over 4},0\right)\) . So, the coordinates of the focus w.r.t old axes are \(\left( {5\over 4},1\right)\)
The equation of the directrix of the parabola y2+4y+4x+2=0 is
- (a)
x=-1
- (b)
x=1
- (c)
x=-3/2
- (d)
x=3/2
We can write the equation of the given parabola as \((y+2)^2=-\left(x-{1\over 2}\right)\)
Shifting the origin (1/2, -2) the equation of parabola becomes Y2=-4X, where X=x-1/2, Y=y+2
The equation of its directrix is X=1
Hence required equation of directrix is \(x-{1\over 2}=1\ or\ x=3/2\)
the eccentricity of conic 9x2+25y2=225 is
- (a)
2/5
- (b)
4/5
- (c)
1/5
- (d)
6/5
The length of the latus rectum of the ellipse 9x2+16y2=144 is
- (a)
4
- (b)
11/4
- (c)
7/2
- (d)
9/2
Given equation of ellipse is 9x 2+16y2 =144
or \({96\over 144}x^2+{16\over144}y^2=1\ or\ {x^2\over (4)^2}+{y^2\over (3)^3}=1\)
∴ a=4, b=3
length of laatus rectum \(={2b^2\over a}={2.9\over 4}={9\over2}\)
the equation of the ellipse whose focus is (1, -1), directrix is the line x-y-3=0 and the eccentricity is \(1\over 2\)is
- (a)
7x2+2xy+7y2-10x+10y+7=0
- (b)
7x2+2xy+7y2+7=0
- (c)
7x2+2xy+7y2-10x-10y-7=0
- (d)
None of these
Let P(x, y) be any point on the ellipse. Then by definition SP=e.PM
\(⇒ \sqrt{(x-1)^2+(y+1)^2}={1\over 2}\left(x-y-3\over \sqrt2\right)\)
⇒ (x-1)2+(y+1)2=1/8(x-y-3)2
⇒ 7x2+2xy+7y2-10x+10y+7=0
If for the ellipse \({x^2\over a^2}+{y^2\over b^2}=1\) y-axis is the minor axis and the length of the latus rectum is one half of the length of its minor axis, then its eccentricity is
- (a)
\(1\over \sqrt 2\)
- (b)
1/2
- (c)
\(\sqrt3\over 2\)
- (d)
3/4
Given that \({2b^2\over a}={1\over 2}.2b⇒{2b^2\over a}=b⇒{b\over a}={1\over 2}\)
\(e={\sqrt{1-{b^2\over a^2}}}=\sqrt{1-{1\over 4}}={\sqrt3\over 2}\)
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
\(2ae=2\times5\times{3\over6}=6\)
Fill in the blanks.
(i) The length of the latus rectum of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) is P.
(ii) The equations of the hyperbola with vertices (±2,0), foci (±3,0) is Q.
(iii) If the distance between the foci of a hyperbola is 16and its eccentricity is 2, then equation of the hyperbola is R.
- (a)
P Q R \(\frac{9}{2}\) \(\frac{y^2}{4}-\frac{x^2}{5}=1\) x2 - 3y2 = 48 - (b)
P Q R \(\frac{3}{2}\) \(\frac{y^2}{4}-\frac{x^2}{5}=1\) 3x2 - y2 = 48 - (c)
P Q R \(\frac{9}{2}\) \(\frac{x^2}{4}-\frac{y^2}{5}=1\) 3x2 - y2 = 48 - (d)
P Q R \(\frac{9}{2}\) \(\frac{x^2}{4}-\frac{y^2}{5}=1\) x2 - y2 = 48
(i) Given equation of hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Now, a2 = 16 ⇒ a = 4 and b2 = 9 ⇒ b=3
Length of latus rectum=\(\frac{2b^2}{a}=\frac{2\times9}{4}=\frac{9}{2}\)
(ii) Vertices are (±2, 0) which lie on x-axis.
So the equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Now, vertices are (±2, 0) ⇒ a = 2
Foci are (±3, 0) ⇒ ae = 3 ⇒ e=\(\frac{3}{2}\)
We know that
b=\(a\sqrt{e^2-1}\Rightarrow b=2\sqrt{\frac{9}{4}-1}=2\frac{\sqrt{5}}{2}=5\)
Thus required equation of hyperbola is
\(\frac{x^2}{(2)^2}-\frac{y^2}{(\sqrt{5})^2}=1\Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=1\)
(iii) Since distance between the foci is 16
⇒ 2ae = 16 ⇒ ae = 8 ⇒ a = 4
Now, b2 = a2(e2 - 1) = (4)2[(2)2- 1] = 16 x 3 = 48
Hence, the equation of hyperbola is \(\frac{x^2}{16}-\frac{y^2}{48}=1\)
⇒ 3x2 - y2 = 48
Parabola is symmetric with respect to the axis of the parabola.
Statement -I: If the equation has a term ,y2then the axis of symmetry is along the X-axis.
Statement-II: If the equation has a term x2, then the axis of symmetry is along the X-axis.
- (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.
If the equation has a y2 term, then the axis of symmetry is along the X-axis and if the equation has an x2 term, then the axis of symmetry is along the Y-axis.
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.