Question

The normal at a variable point P on an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 of eccentricity 'e' meets the axes of the ellipse in Q and R then the locus of the mid-point of QR is a conic with an eccentricity e' such that –

• A. e' = 1
• B. e' = e
• C. e' = 1/e
• D. e' is independent of e

Answer 1

Answer: (B)

e' = e

Answer 2

Equation of normal at P ŗ (a cos q, b sin q)

is $\frac{ax}{\cos\theta}$– $\frac{by}{\sin\theta}$= a² – b²

Q ŗ $\left( \frac{a^{2} - b^{2}}{a}\cos\theta,0 \right)$ R ŗ $\left( 0, - \left( \frac{a^{2} - b^{2}}{b} \right)\sin\theta \right)$

\ M = $\left( \frac{a^{2} - b^{2}}{2a}\cos\theta, - \left( \frac{a^{2} - b^{2}}{2b} \right)\sin\theta \right)$ŗ (h, k)

Hence locus of M

$\frac{h^{2}}{\frac{(a^{2} - b^{2})^{2}}{4a^{2}}}$+ $\frac{k^{2}}{\frac{(a^{2} - b^{2})^{2}}{4b^{2}}}$= 1

which is an ellipse with eccentricity e¢

Q a > b Ž 4a² > 4b²

Ž $\frac{1}{4a^{2}}$ < $\frac{1}{4b^{2}}$Ž $\frac{(a^{2} - b^{2})^{2}}{4a^{2}}$< $\frac{(a^{2} - b^{2})^{2}}{4b^{2}}$

$\frac{(a^{2} - b^{2})^{2}}{4a^{2}}$=$\frac{(a^{2} - b^{2})^{2}}{4b^{2}}$ (1 – e¢²) Ž b² = a²(1 – e¢²)

Ž e = e¢