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]

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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]

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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}$$\frac{b^{2}}{a^{2}}=\frac{3}{5} …..(2)solve (1) \&(2) \mathrm{a}^{2}=\frac{32}{3}, \mathrm{b}^{2}=\frac{32}{5} 0 Q. An ellipse is drawn by taking a diameter of the circle (x-1)^{2}+y^{2}=1 as its semi-minor axis and a diameter of the circle x^{2}+(y-2)^{2}=4 as its semi-major axis. If the centre ofthe ellipse is at the origin and its axes are the coordinate axes, then the equation of theellipse is :(1) x^{2}+4 y^{2}=16(2) 4 x^{2}+y^{2}=4(3) x^{2}+4 y^{2}=8(4) 4 x^{2}+y^{2}=8 [AIEEE-2012] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)Let the equation of ellipse be\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1from the given conditions\mathrm{a}=4 and \mathrm{b}=2$$\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 theellipse $2 \mathrm{x}^{2}+\mathrm{y}^{2}=4$ 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]

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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$$\therefore \mathrm{m}=\pm 2 Q. The equation of the circle passing through the foci of the ellipse \frac{x^{2}}{16}+\frac{y^{2}}{9}=1 and having centre at (0,3) is :(1) x^{2}+y^{2}-6 y-7=0(2) x^{2}+y^{2}-6 y+7=0(3) x^{2}+y^{2}-6 y-5=0(4) x^{2}+y^{2}-6 y+5=0 [JEE (Main)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$$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)$$\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{y}+\mathrm{c}=0passes through (\sqrt{7}, 0)$$7+0-0+\mathrm{c}=0$$\mathrm{c}=-7So \mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{y}-7=0 Q. If a and c are positive real number and the ellipse \frac{x^{2}}{4 c^{2}}+\frac{y^{2}}{c^{2}}=1 has four distinct points in common with the circle x^{2}+y^{2}=9 a^{2}, then(1) 6 \mathrm{ac}+9 \mathrm{a}^{2}-2 \mathrm{c}^{2}>0(2) 6 a c+9 a^{2}-2 c^{2}<0(3) 9 \mathrm{ac}-9 \mathrm{a}^{2}-2 \mathrm{c}^{2}<0(4) 9 \mathrm{ac}-9 \mathrm{a}^{2}-2 \mathrm{c}^{2}>0 [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\mathrm{c}<3 \mathrm{a}<2 \mathrm{c}$$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$$9 \mathrm{ac}-9 \mathrm{a}^{2}-2 \mathrm{c}^{2}>0 Q. Equation of the line passing through the points of intersection of the parabola x^{2}=8 y and the ellipse \frac{x^{2}}{3}+y^{2}=1 is : –(1) y + 3 = 0(2) 3y + 1 = 0(3) 3y – 1 = 0(4) y – 3 = 0 [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Put x^{2}=8 y in \frac{x^{2}}{3}+y^{2}=1$$\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]

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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$$=\mathrm{b}^{2}=4 \quad \Rightarrow \mathrm{b}=2length of minor axis = 4 Q. If the curves \frac{x^{2}}{\alpha}+\frac{y^{2}}{4}=1 and y^{3}=16 x intersect at right angles, then a value of \alpha is :(1) \frac{4}{3}(2) \frac{3}{4}(3) \frac{1}{2}(4) 2 [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\frac{\mathrm{x}^{2}}{\alpha}+\frac{\mathrm{y}^{2}}{4}=1 \& \mathrm{y}^{3}=16 \mathrm{x}d.w.r. to x.\frac{2 \mathrm{x}}{\alpha}+\frac{\mathrm{y}}{2} \mathrm{y}^{\prime}=0 \quad \& \quad 3 \mathrm{y}^{2} \mathrm{y}^{\prime}=16$$\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$$\alpha=\frac{4}{3} \quad as \left(16 \mathrm{x}=\mathrm{y}^{3}\right) Q. A point on the ellipse, 4 \mathrm{x}^{2}+9 \mathrm{y}^{2}=36, where the normal is parallel to the line, 4 \mathrm{x}-2 \mathrm{y}-5 =0, is : -(1) \left(\frac{8}{5},-\frac{9}{5}\right)(2) \left(-\frac{9}{5}, \frac{8}{5}\right)( 3)\left(\frac{8}{5}, \frac{9}{5}\right)( 4)\left(\frac{9}{5}, \frac{8}{5}\right) [JEE-Main (On line)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1Any point (3 \cos \theta, 2 \mathrm{sin} \theta)Slope of normal =2Slope of tangent =-\frac{1}{2}=-\frac{2 \cos \theta}{3 \sin \theta}$$\tan \theta=\frac{4}{3}$$\sin \theta=\frac{4}{5} \quad \cos \theta=\frac{3}{5}Point \left(\frac{9}{5}, \frac{8}{5}\right) Q. The locus of the foot of perpendicular drawn from the centre of the ellipse x^{2}+3 y^{2}=6 on any tangent to it is :-(1) \left(x^{2}-y^{2}\right)^{2}=6 x^{2}+2 y^{2}(2) \left(x^{2}-y^{2}\right)^{2}=6 x^{2}-2 y^{2}(3) \left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}(4) \left(x^{2}+y^{2}\right)^{2}=6 x^{2}-2 y^{2} [JEE(Main)-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)Let the foot of perpendicular be (h, k)then \mathrm{m}_{\mathrm{op}}=\frac{\mathrm{k}}{\mathrm{h}}equation of tangent is \mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}$$\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}}$$\left(\mathrm{k}+\frac{\mathrm{h}^{2}}{\mathrm{k}}\right)^{2}=\frac{6 \mathrm{h}^{2}}{\mathrm{k}^{2}}+2multiply by \mathrm{k}^{2}$$\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 thelatera 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]

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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]

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Sol. (3)Eccentricity of ellipse $=\frac{1}{2}$

• April 23, 2022 at 8:34 pm

Potato has been here

• February 17, 2021 at 2:52 pm

Thank you

• February 13, 2021 at 5:15 pm

No questions after 2017 and here I am in 2021?

• December 28, 2020 at 7:02 am

Thanks a lot 😊.

• October 12, 2020 at 11:54 pm

👍👍👍👍

• October 25, 2020 at 10:21 am

Nice👍👍

• August 16, 2020 at 8:10 am

Plz try to update the questions from recent years too. Good questions. Thank you

• August 10, 2020 at 8:05 pm

Thanks

• August 6, 2020 at 4:16 pm

update latest quesations

• July 23, 2020 at 8:59 pm

Nice Questions

• July 19, 2020 at 12:32 am

Good

• July 16, 2020 at 11:47 pm

I want to see 2018,2019 & 2020 questions also since the pattern has been changed with the year.plz…. update as fast as possible .

• 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

• 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

• December 22, 2020 at 10:01 pm

u are absolutely correct jayasree