Question

The length of the diameter of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ perpendicular to the asymptotes of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ passing through the first and third quadrant is

• A. $\frac{100}{\sqrt{431}}$
• B. $\frac{150}{\sqrt{481}}$
• C. $\frac{25}{\sqrt{3}}$
• D. $11\sqrt{2}$

Answer 1

Answer: (B)

$\frac{150}{\sqrt{481}}$

Answer 2

$\frac{x^{2}}{16}–\frac{y^{2}}{9} = 1$

y = 3/4 x [equation of asymptotes]

line ⊥to asymptotes and passing through centre (diameter)

y = $- \frac{4}{3}x$

point of intersection of line (diameter)

y = $- \frac{4}{3}x$and ellipse $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

A = $\left( \frac{45}{\sqrt{481}}, - \frac{60}{\sqrt{481}} \right)B = \left( - \frac{45}{\sqrt{481}},\frac{60}{\sqrt{481}} \right)$

Diameter = ${\sqrt{\left( \frac{90}{\sqrt{481}} \right)^{2} + \left( \frac{120}{\sqrt{481}} \right)^{2}}}^{} = \frac{150}{\sqrt{481}}$