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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 10−5\sqrt{10}-\sqrt5. The equation of the ellipse is:
Ax236+y218=1\dfrac{x^2}{36}+\dfrac{y^2}{18}=1
Bx240+y220=1\dfrac{x^2}{40}+\dfrac{y^2}{20}=1
Cx220+y210=1\dfrac{x^2}{20}+\dfrac{y^2}{10}=1
Dx210+y25=1\dfrac{x^2}{10}+\dfrac{y^2}{5}=1correct
Solution
Step 1: The ends of the minor axis are (0,±b)(0,\pm b) and the focus is (ae,0)(ae,0). The lines from (0,b)(0,b) and (0,−b)(0,-b) to (ae,0)(ae,0) have slopes −bae-\dfrac{b}{ae} and bae\dfrac{b}{ae}; perpendicularity gives
−b2a2e2=−1  ⇒  b2=a2e2=a2−b2  ⇒  a2=2b2.-\frac{b^2}{a^2e^2}=-1 \;\Rightarrow\; b^2=a^2e^2=a^2-b^2 \;\Rightarrow\; a^2=2b^2.
 Hence e2=a2−b2a2=12e^2=\dfrac{a^2-b^2}{a^2}=\dfrac12, so e=12e=\dfrac{1}{\sqrt2}. Step 2: The focus-to-nearer-vertex distance is a−ae=a(1−12)a-ae=a\left(1-\dfrac{1}{\sqrt2}\right). Set it equal to 10−5=5(2−1)\sqrt{10}-\sqrt5=\sqrt5(\sqrt2-1):
a⋅2−12=5(2−1)  ⇒  a2=5  ⇒  a=10,a2=10.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 b2=a22=5b^2=\dfrac{a^2}{2}=5, giving x210+y25=1\dfrac{x^2}{10}+\dfrac{y^2}{5}=1. Correct answer: (4)
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