Question

Length of subtangent and subnormal at the point $\left( \frac{- 5\sqrt{3}}{2},2 \right)$of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$are

• A. $\left( \frac{5\sqrt{3}}{2} - \frac{10}{\sqrt{3}} \right)$, $\frac{8\sqrt{3}}{5}$
• B. $\left( \frac{5\sqrt{3}}{2} + \frac{10}{\sqrt{3}} \right)$, $\frac{8\sqrt{3}}{10}$
• C. $\left( \frac{5\sqrt{3}}{2} + \frac{12}{\sqrt{3}} \right)$, $\frac{16\sqrt{3}}{5}$
• D. None of these

Answer 1

Answer: (A)

$\left( \frac{5\sqrt{3}}{2} - \frac{10}{\sqrt{3}} \right)$, $\frac{8\sqrt{3}}{5}$

Answer 2

Here $a^{2} = 25,b^{2} = 16,x_{1} = \frac{- 5\sqrt{3}}{2}$. Length of subtangent =$\left| \frac{a^{2}}{x_{1}} - x_{1} \right| = \left| \frac{25}{- 5\sqrt{3}/2} + \frac{5\sqrt{3}}{2} \right| = \left| \frac{5\sqrt{3}}{2} - \frac{10}{\sqrt{3}} \right|$.

Length of subnormal $= \left| \frac{b^{2}}{a^{2}}x_{1} \right| = \left| \frac{16}{25}\left( \frac{- 5\sqrt{3}}{2} \right) \right| = \left| \frac{8\sqrt{3}}{5} \right|$