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Question: If normal to ellipse \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\) at \(\left( ae,\frac{b^{2}}{a...

If normal to ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (ae,b2a)\left( ae,\frac{b^{2}}{a} \right) is passing through (0, -2b), then value of eccentricity is

A

2−1\sqrt{2} - 1

B

2(2−1)\left( \sqrt{2} - 1 \right)

C

2(2−1)\sqrt{2\left( \sqrt{2} - 1 \right)}

D

None of these

Answer

2(2−1)\sqrt{2\left( \sqrt{2} - 1 \right)}

Explanation

Solution

Equation of the normal at (ae,b2a)\left( ae,\frac{b^{2}}{a} \right) is

y1b2(x−x1)−x1a2(y−y1)\frac{y_{1}}{b^{2}}\left( x - x_{1} \right) - \frac{x_{1}}{a^{2}}\left( y - y_{1} \right)=0

⇒ 1a(x−ae)−ea(y−b2a)=0\frac{1}{a}(x - ae) - \frac{e}{a}\left( y - \frac{b^{2}}{a} \right) = 0∴ (0, -2b) is lying on (1)

∴ -e + eab(2+ba)\frac{e}{a}b\left( 2 + \frac{b}{a} \right) = 0 ⇒ b(2a+b) = a2

⇒ 2ab = a2-b2 ⇒ 2ab = a2e2

⇒ 4a2(1−e2)=a2e44a^{2}\left( 1 - e^{2} \right) = a^{2}e^{4}

⇒ e4+4e2−4=0⇒e2=−4±322e^{4} + 4e^{2} - 4 = 0 \Rightarrow e^{2} = \frac{- 4 \pm \sqrt{32}}{2}∴ e = 2(2−1)\sqrt{2\left( \sqrt{2} - 1 \right)}