Question

$L_1$ is a tangent drawn to the curve $x^2-4 y^2=16$ at $A\left(5, \frac{3}{2}\right) \cdot L_2$ is another tangent parallel to $L_1$ which meets the curve at $B$. $L_3$ and $L_4$ are normals to the curve at $A$ and $B$. Lines $L_1, L_2, L_3, L_4$ forms a rectangle, then

• A. Equation of tangent at $B$ is $6 y=5 x-16$.
• B. Equation of normal at $B$ is $12 x+10 y+75=0$
• C. Radius of largest circle inscribed in the rectangle is $\frac{32}{\sqrt{61}}$.
• D. Radius of the circle circumscribing the rectangle is $\frac{\sqrt{109}}{2}$

Answer 1

Answer: (B), (D)

$\frac{x^2}{16}-\frac{y^2}{4}=1$
$\left(5, \frac{3}{2}\right)$ lies on the curve.
tangent at $A$
$\frac{5 x}{16}-\frac{3}{2}\left(\frac{y}{4}\right)=1 \Rightarrow 5 x-6 y=16$
tangent at B
$5 x-6 y=-6$
Normal at A
$6 x+5 y=\lambda $
$A\left(5, \frac{3}{2}\right) $
$30+\frac{15}{2}=1 \Rightarrow \lambda=\frac{75}{2} $
$6 x+5 y=\frac{75}{2}$
$\|\left.\right|^{ ly }$ normal at $B$ is $6 x+5 y=\frac{-75}{2}$
distance between tangents is $2 r =\frac{32}{\sqrt{25+36}}$
$\therefore r =\frac{16}{\sqrt{61}}$
Radius of circumcircle of rectangle is $R=\sqrt{25+\frac{9}{4}}=\frac{\sqrt{109}}{2}$