Question

If the normal at an end of a latus rectum passes through the extremity of the minor axis then eccentricity of the ellipse is given by

• A. e² + e + 1 = 0
• B. e⁴ + e² + 1 = 0
• C. e⁴ – e² – 1 = 0
• D. e⁴ + e² – 1 = 0

Answer 1

Answer: (D)

e⁴ + e² – 1 = 0

Answer 2

Let equation of ellipse be

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and L$\left( ae,\mspace{6mu}\frac{b^{2}}{a} \right)$

be one end of a latus rectum through S(ae, 0)

Now equation of normal at L is $\frac{a^{2}x}{ae} - \frac{b^{2}y}{b^{2}/a} = a^{2} - b^{2}$⇒ $\frac{ax}{e} - ay = a^{2}e^{2}$

⇒ $\frac{x}{e} - y = ae^{2}$

Now this normal passing through L’(0, b)

∴ b = ae²

⇒ b² =a²e⁴

∴ a²(1 – e²) = a²e⁴

⇒ e⁴ + e² – 1 = 0