Question

Find the length of sub normal of $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $(3,0)$

• A. $\frac{3}{4}$
• B. $\frac{13}{4}$
• C. $\frac{12}{9}$
• D. None of these

Answer 1

The correct option is C $\frac{12}{9}$

Length of subnormal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(x_1,y_1)=(1-e^2)x_1$

for ellipse, $\frac{x^2}{9}+\frac{y^2}{4}=1$

$a^2=9,b^2=4,-(1-e^2)=\frac{b^2}{a^2}\Rightarrow \frac{4}{9}$

length of subnormal to $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $(3,0)\Rightarrow (1-e^2)x_1$

$\Rightarrow \frac{4}{9}\times 3$

$\Rightarrow \frac{12}{9}$