JEE Main 2021 — Mathematics Coordinate Geometry

2021mcqmedium

Let ${E}_{1}:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>b.$ Let ${E}_{2}$ be another ellipse such that it touches the end points of major axis of ${E}_{1}$ and the foci of ${E}_{2}$ are the end points of minor axis of ${E}_{1}.$ If ${E}_{1}$ and ${E}_{2}$ have same eccentricities, then its value is:

Official previous-year question

Held on 22 Jul 2021 · Verified 6 Jul 2026.

Options

A
$\frac{-1+\sqrt{5}}{2}$
B
$\frac{-1+\sqrt{8}}{2}$
C
$\frac{-1+\sqrt{3}}{2}$
D
$\frac{-1+\sqrt{6}}{2}$

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Solution

Given ellipse is ${E}_{1}:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>b$ and ${E}_{2}$ is another ellipse which touches the end points of the major axis of ${E}_{1}$ and foci of ${E}_{2}$ are at the minor axis of ${E}_{1},$ thus the two ellipses are given by the following diagram.

JEE Main 2021 Mathematics, Ellipse — solution figure

Since, the ellipse ${E}_{2}$ touches the major axis of the ellipse ${E}_{1}$ , hence, the minor axis of the ellipse ${E}_{2}$ is ${a}_{1}=a.$

Now, let the major axis of the ellipse ${E}_{2}$ is ${b}_{1}=c$ and it is obvious from the diagram and the given conditions that ${b}_{1}>{a}_{1}.$

We know that the eccentricity of an ellipse $\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1,A>B$ is $\sqrt{1-\frac{{B}^{2}}{{A}^{2}}}.$

Thus, for the ellipse ${E}_{1},$ we have $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$ and for the ellipse ${E}_{2},$ we have ${e}_{1}=\sqrt{1-\frac{{a}^{2}}{{c}^{2}}}.$

Given, $e={e}_{1}$

$\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\frac{{a}^{2}}{{c}^{2}}}$

$\Rightarrow 1-\frac{{b}^{2}}{{a}^{2}}=1-\frac{{a}^{2}}{{c}^{2}}$

$\Rightarrow \frac{{b}^{2}}{{a}^{2}}=\frac{{a}^{2}}{{c}^{2}}$

$\Rightarrow {c}^{2}=\frac{{a}^{4}}{{b}^{2}}$

$\Rightarrow c=\frac{{a}^{2}}{b}...(i)$

Also, given that the foci of ${E}_{2}$ are the end points of the minor axis of ${E}_{1},$ thus $b=ce$

$\Rightarrow c=\frac{b}{e}...(ii)$

From the above two equations, we get $\frac{b}{e}=\frac{{a}^{2}}{b}$

$\Rightarrow e=\frac{{b}^{2}}{{a}^{2}}$

Now, using the definition of eccentricity, we get $e=\frac{{b}^{2}}{{a}^{2}}=1-{e}^{2}$

$\Rightarrow {e}^{2}+e-1=0$

Now, applying the Sridharacharya's formula for the roots of a quadratic equation, i.e. if $a{x}^{2}+bx+c=0,a\neq 0,$ then $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a},$ we get

$e=\frac{-1\pm \sqrt{{1}^{2}-4\times 1\times (-1)}}{2\times 1}$

$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}$

$\Rightarrow e=\frac{-1\pm \sqrt{5}}{2}$

But, eccentricity can never be negative, hence $e=\frac{-1+\sqrt{5}}{2}.$

Your note

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Solution

JEE Main 2021 Mathematics, Ellipse — solution figure

Since, the ellipse ${E}_{2}$ touches the major axis of the ellipse ${E}_{1}$ , hence, the minor axis of the ellipse ${E}_{2}$ is ${a}_{1}=a.$

Now, let the major axis of the ellipse ${E}_{2}$ is ${b}_{1}=c$ and it is obvious from the diagram and the given conditions that ${b}_{1}>{a}_{1}.$

We know that the eccentricity of an ellipse $\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1,A>B$ is $\sqrt{1-\frac{{B}^{2}}{{A}^{2}}}.$

Thus, for the ellipse ${E}_{1},$ we have $e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$ and for the ellipse ${E}_{2},$ we have ${e}_{1}=\sqrt{1-\frac{{a}^{2}}{{c}^{2}}}.$

Given, $e={e}_{1}$

$\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\frac{{a}^{2}}{{c}^{2}}}$

$\Rightarrow 1-\frac{{b}^{2}}{{a}^{2}}=1-\frac{{a}^{2}}{{c}^{2}}$

$\Rightarrow \frac{{b}^{2}}{{a}^{2}}=\frac{{a}^{2}}{{c}^{2}}$

$\Rightarrow {c}^{2}=\frac{{a}^{4}}{{b}^{2}}$

$\Rightarrow c=\frac{{a}^{2}}{b}...(i)$

Also, given that the foci of ${E}_{2}$ are the end points of the minor axis of ${E}_{1},$ thus $b=ce$

$\Rightarrow c=\frac{b}{e}...(ii)$

From the above two equations, we get $\frac{b}{e}=\frac{{a}^{2}}{b}$

$\Rightarrow e=\frac{{b}^{2}}{{a}^{2}}$

Now, using the definition of eccentricity, we get $e=\frac{{b}^{2}}{{a}^{2}}=1-{e}^{2}$

$\Rightarrow {e}^{2}+e-1=0$

Now, applying the Sridharacharya's formula for the roots of a quadratic equation, i.e. if $a{x}^{2}+bx+c=0,a\neq 0,$ then $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a},$ we get

$e=\frac{-1\pm \sqrt{{1}^{2}-4\times 1\times (-1)}}{2\times 1}$

$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}$

$\Rightarrow e=\frac{-1\pm \sqrt{5}}{2}$

But, eccentricity can never be negative, hence $e=\frac{-1+\sqrt{5}}{2}.$

JEE Main 2021 — Mathematics Coordinate Geometry

Options

Did you get this right?

Solution

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JEE Main Coordinate Geometry in other years

More JEE Main Coordinate Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Vectors & 3D Geometry

Probability & Statistics

Trigonometry

Keep practicing Coordinate Geometry

JEE Main 2021 — Mathematics Coordinate Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Coordinate Geometry in other years

More JEE Main Coordinate Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Vectors & 3D Geometry

Probability & Statistics

Trigonometry

Keep practicing Coordinate Geometry