Question

Find the length of normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(3,5)$.

• A. $\frac{9}{5}$
• B. $\frac{5}{9}$
• C. $\frac{\sqrt{34}}{9}$
• D. $\frac{5\sqrt{34}}{9}$

Answer 1

The correct option is D $\frac{5\sqrt{34}}{9}$

Length of normal can be found using the formula

$y\times \sqrt{1+(dy/dx{)}^2}$

Given $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{2x}{a^2}+\frac{2y}{b^2}\times \frac{dy}{dx}=0$

$\frac{dy}{dx}\times \frac{2y}{b^2}=-\frac{-2x}{a^2}$

$\frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}$

Length of the normal is

$y\times \sqrt{(}1+(dy/dx{)}^2$=$y\sqrt{1+(\frac{-x}{y}{)}^2}$

=$5\times \sqrt{1+\frac{25}{9}}$$=5\sqrt{\frac{34}{9}}$

Option D is the correct answer