Question

From the point $A(4,3)$ tangents are drawn to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ to touch the ellipse at $B$ and $C$. EF is a tangent to the ellipse parallel to the line $BC$ and towards the point $A$. The distance of $A$ from EF is equal to

• A. $\frac{4 \sqrt{18}-12}{5} $
• B. $\frac{24-4 \sqrt{18}}{5}$
• C. $\frac{12-2 \sqrt{18}}{5}$
• D. $\frac{12+2 \sqrt{18}}{5}$

Answer 1

Answer: (B)

Equation of chord of contact at $A(4,3)$
$\frac{x}{4}+\frac{y}{3}=1$
Slope of line EF is $\frac{-3}{4}$
Equation of $EF$, (EF is tangent of ellipse)
$y=m x+\sqrt{a^2 m^2+b^2} $
$y=\frac{-3}{4} x+\sqrt{16 \cdot \frac{9}{16}+9} $
$y=\frac{-3}{4} x+\sqrt{18} $
$\text { EF, } 3 x+4 y-4 \sqrt{18}=0$
$d =\left|\frac{12+12-4 \sqrt{18}}{5}\right|=\left|\frac{24-4 \sqrt{18}}{5}\right|$