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Ellipse from Minor-Axis Right Angle at Focus | JEE

JEE Maths question with a full step-by-step solution.

Question

In the standard ellipse, the lines joining the ends of the minor axis to one focus are at right angles. The distance between the focus and the nearer vertex is

\sqrt{10}-\sqrt5

. The equation of the ellipse is:

\dfrac{x^2}{36}+\dfrac{y^2}{18}=1

\dfrac{x^2}{40}+\dfrac{y^2}{20}=1

\dfrac{x^2}{20}+\dfrac{y^2}{10}=1

\dfrac{x^2}{10}+\dfrac{y^2}{5}=1

correct

Solution

Step 1: The ends of the minor axis are

(0,\pm b)

and the focus is

(ae,0)

. The lines from

(0,b)

and

(0,-b)

(ae,0)

have slopes

-\dfrac{b}{ae}

and

\dfrac{b}{ae}

; perpendicularity gives

-\frac{b^2}{a^2e^2}=-1 \;\Rightarrow\; b^2=a^2e^2=a^2-b^2 \;\Rightarrow\; a^2=2b^2.

Hence

e^2=\dfrac{a^2-b^2}{a^2}=\dfrac12

, so

e=\dfrac{1}{\sqrt2}

. Step 2: The focus-to-nearer-vertex distance is

a-ae=a\left(1-\dfrac{1}{\sqrt2}\right)

. Set it equal to

\sqrt{10}-\sqrt5=\sqrt5(\sqrt2-1)

a\cdot\frac{\sqrt2-1}{\sqrt2}=\sqrt5(\sqrt2-1) \;\Rightarrow\; \frac{a}{\sqrt2}=\sqrt5 \;\Rightarrow\; a=\sqrt{10},\quad a^2=10.

Step 3: Then

b^2=\dfrac{a^2}{2}=5

, giving

\dfrac{x^2}{10}+\dfrac{y^2}{5}=1

. Correct answer: (4)

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