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Q. The ellipse $x^{2}+4 y^{2}=4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is :-
(1) $4 x^{2}+48 y^{2}=48$
(2) $4 x^{2}+64 y^{2}=48$
(3) $x^{2}+16 y^{2}=16$
(4) $x^{2}+12 y^{2}=16$
[AIEEE-2009]
Ans. (4)
Q. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point (–3, 1) and has eccentricity $\sqrt{2 / 5}$ is :-
(1) $3 x^{2}+5 y^{2}-15=0$
(2) $5 x^{2}+3 y^{2}-32=0$
(3) $3 x^{2}+5 y^{2}-32=0$
(4) $5 x^{2}+3 y^{2}-48=0$
[AIEEE-2011]
Ans. (3)
$\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \mathrm{e}=\sqrt{\frac{2}{5}}$
It passes through $(-3,1)$ so $\frac{9}{a^{2}}+\frac{1}{b^{2}}=1$ .....(1)
also $1-\frac{b^{2}}{a^{2}}=e^{2} \Rightarrow 1-\frac{b^{2}}{a^{2}}=\frac{2}{5}$
$\frac{b^{2}}{a^{2}}=\frac{3}{5}$ .....(2)
solve $(1) \&(2) \mathrm{a}^{2}=\frac{32}{3}, \mathrm{b}^{2}=\frac{32}{5} 0$
Q. An ellipse is drawn by taking a diameter of the circle $(x-1)^{2}+y^{2}=1$ as its semi-minor axis and a diameter of the circle $x^{2}+(y-2)^{2}=4$ as its semi-major axis. If the centre of
the ellipse is at the origin and its axes are the coordinate axes, then the equation of the
ellipse is :
(1) $x^{2}+4 y^{2}=16$
(2) $4 x^{2}+y^{2}=4$
(3) $x^{2}+4 y^{2}=8$
(4) $4 x^{2}+y^{2}=8$
[AIEEE-2012]
Ans. (1)
Let the equation of ellipse be
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
from the given conditions
$\mathrm{a}=4$ and $\mathrm{b}=2$
$\therefore \mathrm{Eq}$ of ellipse is $\frac{\mathrm{x}^{2}}{16}+\frac{\mathrm{y}^{2}}{4}=1$
or $\mathrm{x}^{2}+4 \mathrm{y}^{2}=16$
Q. Statement-1: An equation of a common tangent to the parabola $\mathrm{y}^{2}=16 \sqrt{3} \mathrm{x}$ and the
ellipse $2 \mathrm{x}^{2}+\mathrm{y}^{2}=4$ is $\mathrm{y}=2 \mathrm{x}+2 \sqrt{3}$
Statement-2: If the line $y=m x+\frac{4 \sqrt{3}}{m},(m \neq 0)$ is a common tangent to the parabola $y^{2}=$ $$
16 \sqrt{3} x \text { and the ellipse } 2 x^{2}+y^{2}=4, \text { then } m \text { satisfies } m^{4}+2 m^{2}=24$$
(1) Statement–1 is true, Statement–2 is false.
(2) Statement–1 is false, Statement–2 is true.
(3) Statement–1 is true, Statement–2 is true ; Statement–2 is a correct explanation for Statement–1.
(4) Statement–1 is true, Statement–2 is true ; Statement–2 is not a correct explanation for Statement– 1.
[AIEEE-2012]
Ans. (3)
Let equation of any tangent to $\mathrm{y}^{2}=16 \sqrt{3} \mathrm{x}$
be $\mathrm{y}=\mathrm{mx}+\frac{4 \sqrt{3}}{\mathrm{m}}$
and equation of any tangent to $2 \mathrm{x}^{2}+\mathrm{y}^{2}=4$
be $\mathrm{y}=\mathrm{mx}+\sqrt{2 \mathrm{m}^{2}+4} \ldots \ldots$ (ii)
but $(\mathrm{i})$ and (ii) are same lines
$\therefore \frac{4 \sqrt{3}}{\mathrm{m}}=\sqrt{2 \mathrm{m}^{2}+4}$
$\Rightarrow \mathrm{m}^{4}+2 \mathrm{m}^{2}-24=0$
$\Rightarrow \mathrm{m}^{2}=-6,4$
$\therefore \mathrm{m}=\pm 2$
Q. The equation of the circle passing through the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and having centre at $(0,3)$ is :
(1) $x^{2}+y^{2}-6 y-7=0$
(2) $x^{2}+y^{2}-6 y+7=0$
(3) $x^{2}+y^{2}-6 y-5=0$
(4) $x^{2}+y^{2}-6 y+5=0$
[JEE (Main)-2013]
Ans. (1)
$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
$e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{\sqrt{7}}{4}$
foci $(\pm a e, 0) \equiv(\pm \sqrt{7}, 0)$
centre of circle is $(0,3)$
$\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{y}+\mathrm{c}=0$
passes through $(\sqrt{7}, 0)$
$7+0-0+\mathrm{c}=0$
$\mathrm{c}=-7$
So $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{y}-7=0$
Q. If a and $c$ are positive real number and the ellipse $\frac{x^{2}}{4 c^{2}}+\frac{y^{2}}{c^{2}}=1$ has four distinct points in common with the circle $x^{2}+y^{2}=9 a^{2},$ then
(1) $6 \mathrm{ac}+9 \mathrm{a}^{2}-2 \mathrm{c}^{2}>0$
(2) $6 a c+9 a^{2}-2 c^{2}<0$
(3) $9 \mathrm{ac}-9 \mathrm{a}^{2}-2 \mathrm{c}^{2}<0$
(4) $9 \mathrm{ac}-9 \mathrm{a}^{2}-2 \mathrm{c}^{2}>0$
[JEE-Main (On line)-2013]
Ans. (4)
$\mathrm{c}<3 \mathrm{a}<2 \mathrm{c}$
$3 \mathrm{a}-\mathrm{c}>0$
$2 \mathrm{c}-3 \mathrm{a}>0$
$\because(3 \mathrm{a}-\mathrm{c})(2 \mathrm{c}-3 \mathrm{a})>0$
$9 \mathrm{ac}-9 \mathrm{a}^{2}-2 \mathrm{c}^{2}>0$
Q. Equation of the line passing through the points of intersection of the parabola $x^{2}=8 y$ and the ellipse $\frac{x^{2}}{3}+y^{2}=1$ is : -
(1) y + 3 = 0
(2) 3y + 1 = 0
(3) 3y – 1 = 0
(4) y – 3 = 0
[JEE-Main (On line)-2013]
Ans. (3)
Put $x^{2}=8 y$ in $\frac{x^{2}}{3}+y^{2}=1$
$\quad=(y+3)(3 y-1)=0$
$\mathrm{y}+3=0$
$3 \mathrm{y}-1=0$
Q. Let the equations of two ellipses be $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}=1$ and $\mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{16}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 .$ If the product of their eccentricities is $\frac{1}{2},$ then the length of the minor axis of ellipse $\mathrm{E}_{2}$ is :-
(1) 9 (2) 8 (3) 2 (4) 4
[JEE-Main (On line)-2013]
Ans. (4)
$\mathrm{e}_{1} \mathrm{e}_{2}=\frac{1}{2}$
$\Rightarrow \quad \sqrt{1-\frac{2}{3}} \sqrt{1-\frac{\mathrm{b}^{2}}{16}}=\frac{1}{2} \quad$ if $16>\mathrm{b}^{2}$
$=\left(16-\mathrm{b}^{2}\right)=12$
$=\mathrm{b}^{2}=4 \quad \Rightarrow \mathrm{b}=2$
length of minor axis = 4
Q. If the curves $\frac{x^{2}}{\alpha}+\frac{y^{2}}{4}=1$ and $y^{3}=16 x$ intersect at right angles, then a value of $\alpha$ is :
(1) $\frac{4}{3}$
(2) $\frac{3}{4}$
(3) $\frac{1}{2}$
(4) 2
[JEE-Main (On line)-2013]
Ans. (1)
$\frac{\mathrm{x}^{2}}{\alpha}+\frac{\mathrm{y}^{2}}{4}=1 \& \mathrm{y}^{3}=16 \mathrm{x}$
d.w.r. to x.
$\frac{2 \mathrm{x}}{\alpha}+\frac{\mathrm{y}}{2} \mathrm{y}^{\prime}=0 \quad \& \quad 3 \mathrm{y}^{2} \mathrm{y}^{\prime}=16$
$\mathrm{y}^{\prime}=-\frac{4 \mathrm{x}}{\alpha \mathrm{y}} \& \mathrm{y}^{\prime}=\frac{16}{3 \mathrm{y}^{2}}$
Both are orthogonal $\Rightarrow-\frac{4 \mathrm{x}}{\alpha \mathrm{y}} \cdot \frac{16}{3 \mathrm{y}^{2}}=-1$
$\alpha=\frac{4}{3} \quad$ as $\left(16 \mathrm{x}=\mathrm{y}^{3}\right)$
Q. A point on the ellipse, $4 \mathrm{x}^{2}+9 \mathrm{y}^{2}=36,$ where the normal is parallel to the line, $4 \mathrm{x}-2 \mathrm{y}-5$ $=0,$ is : $-$
(1) $\left(\frac{8}{5},-\frac{9}{5}\right)$
(2) $\left(-\frac{9}{5}, \frac{8}{5}\right)$
( 3)$\left(\frac{8}{5}, \frac{9}{5}\right)$
( 4)$\left(\frac{9}{5}, \frac{8}{5}\right)$
[JEE-Main (On line)-2013]
Ans. (4)
$\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$
Any point $(3 \cos \theta, 2 \mathrm{sin} \theta)$
Slope of normal $=2$
Slope of tangent $=-\frac{1}{2}=-\frac{2 \cos \theta}{3 \sin \theta}$
$\tan \theta=\frac{4}{3}$
$\sin \theta=\frac{4}{5} \quad \cos \theta=\frac{3}{5}$
Point $\left(\frac{9}{5}, \frac{8}{5}\right)$
Q. The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2}+3 y^{2}=6$ on any tangent to it is :-
(1) $\left(x^{2}-y^{2}\right)^{2}=6 x^{2}+2 y^{2}$
(2) $\left(x^{2}-y^{2}\right)^{2}=6 x^{2}-2 y^{2}$
(3) $\left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}$
(4) $\left(x^{2}+y^{2}\right)^{2}=6 x^{2}-2 y^{2}$
[JEE(Main)-2014]
Ans. (3)
Let the foot of perpendicular be (h, k)
then $\mathrm{m}_{\mathrm{op}}=\frac{\mathrm{k}}{\mathrm{h}}$
equation of tangent is $\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}$
$\mathrm{y}=\mathrm{mx} \pm \sqrt{6 \mathrm{m}^{2}+2}$
satisfied by $(\mathrm{h}, \mathrm{k})$ and $\mathrm{m}=-\frac{1}{\mathrm{m}_{\mathrm{op}}}=-\frac{\mathrm{h}}{\mathrm{k}}$
$\left(\mathrm{k}+\frac{\mathrm{h}^{2}}{\mathrm{k}}\right)^{2}=\frac{6 \mathrm{h}^{2}}{\mathrm{k}^{2}}+2$
multiply by $\mathrm{k}^{2}$
$\left(\mathrm{k}^{2}+\mathrm{h}^{2}\right)^{2}=6 \mathrm{h}^{2}+2 \mathrm{k}^{2}$
$\Rightarrow\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=6 \mathrm{x}^{2}+2 \mathrm{y}^{2}$
Let the foot of perpendicular be (h, k)
then $\mathrm{m}_{\mathrm{op}}=\frac{\mathrm{k}}{\mathrm{h}}$
equation of tangent is $\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}$
$\mathrm{y}=\mathrm{mx} \pm \sqrt{6 \mathrm{m}^{2}+2}$
satisfied by $(\mathrm{h}, \mathrm{k})$ and $\mathrm{m}=-\frac{1}{\mathrm{m}_{\mathrm{op}}}=-\frac{\mathrm{h}}{\mathrm{k}}$
$\left(\mathrm{k}+\frac{\mathrm{h}^{2}}{\mathrm{k}}\right)^{2}=\frac{6 \mathrm{h}^{2}}{\mathrm{k}^{2}}+2$
multiply by $\mathrm{k}^{2}$
$\left(\mathrm{k}^{2}+\mathrm{h}^{2}\right)^{2}=6 \mathrm{h}^{2}+2 \mathrm{k}^{2}$
$\Rightarrow\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=6 \mathrm{x}^{2}+2 \mathrm{y}^{2}$
Q. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ is :
(1) $\frac{27}{2}$
(2) 27
(3) $\frac{27}{4}$
(4) 18
[JEE(Main)-2015]
Ans. (2)
Q. The eccentricity of an ellipse centre is at the origin is $\frac{1}{2} .$ If one of its directices is $\mathrm{x}=-4,$ then the equation of the normal to it at $\left(1, \frac{3}{2}\right)$ is : -
(1) x + 2y = 4
(2) 2y – x = 2
(3) 4x – 2y = 1
(4) 4x + 2y = 7
[JEE(Main)-2017]
Ans. (3)
Eccentricity of ellipse $=\frac{1}{2}$
Comments
Suggestion
Aug. 16, 2020, 8:10 a.m.
Plz try to update the questions from recent years too. Good questions. Thank you
Abc
July 16, 2020, 11:47 p.m.
I want to see 2018,2019 & 2020 questions also since the pattern has been changed with the year.plz.... update as fast as possible .