Ellipse – JEE Main Previous Year Question with Solutions

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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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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 $\mathrm{k}^{2}$

$\left(\mathrm{k}^{2}+\mathrm{h}^{2}\right)^{2}=6 \mathrm{h}^{2}+2 \mathrm{k}^{2}$

$\Rightarrow\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=6 \mathrm{x}^{2}+2 \mathrm{y}^{2}$


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]

Sol. (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]

Sol. (3)

Eccentricity of ellipse $=\frac{1}{2}$


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Comments
  • July 6, 2020 at 12:46 pm

    Good questions and quality of questions has been increasing from year to year

  • June 23, 2020 at 5:46 pm

    yup good…

  • June 3, 2020 at 8:10 am

    What did Jayasree say it is correct.

  • May 22, 2020 at 11:42 am

    Chill bro

  • Raj
    May 16, 2020 at 3:09 pm

    Plz give all the previous years questions related to some chapter

  • April 10, 2020 at 2:00 pm

    It would be very better if there are questions from remaining years also