Question

Normal to the ellipse $\frac{x^{2}}{84} + \frac{y^{2}}{49} = 1$ intersects the major and minor axis at P and Q respectively then locus of the point dividing segment PQ in 2 : 1 is –

• A. $\frac{64x^{2}}{25} + \frac{49y^{2}}{100} = 1$
• B. $\frac{64x^{2}}{100} + \frac{49y^{2}}{25} = 1$
• C. 64x² + 49y² = 225
• D. $\frac{49x^{2}}{100} + \frac{64y^{2}}{25} = 1$

Answer 1

Answer: (A)

$\frac{64x^{2}}{25} + \frac{49y^{2}}{100} = 1$

Answer 2

Equation of normal 8x secq – 7y cosecq = 15

$P\left( \frac{15}{8}\cos\theta,0 \right)$, Q $\left( 0,\frac{- 15}{7}\sin\theta \right)$

3h = $\frac{15}{8}\cos\theta$, 3k = $- \frac{30}{7}\sin\theta$

cosq = $\frac{8h}{5}$, sinq = $\frac{- 7k}{10}$

hence locus $\frac{64x^{2}}{25} + \frac{49y^{2}}{100}$ = 1