Question

Find the length of sub-normal of $\frac{x^2}{25}-\frac{y^2}{16}=1$. at $(5,0)$.

• A. $\frac{23}{16}$
• B. $\frac{22}{16}$
• C. $\frac{16}{5}$
• D. None of these

Answer 1

Answer: (C)

$\frac{16}{5}$

Length of sub-normal
$=\left|y_1{\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}\right|$

$\frac{dy}{dx}=\frac{16x}{25y}$

put this in length of sub-normal equation

$=\left|y_1{\left(\frac{16x_1}{25y_1}\right)}_{(5,0)}\right|$

$=\frac{80}{25}$

$=\frac{16}{5}$