Question

A variable line drawn through the point of intersection of the lines $\frac{x}{4}+\frac{y}{8}=1$ and $\frac{x}{8}+\frac{y}{4}=1$ and meets the coordinate axes at $A$ and $B$. Locus of mid point of $AB$ is $C$ then

• A. $C$ will be a hyperbola with eccentricity of $\sqrt{2}$
• B. $C$ will be a hyperbola whose asymptote are $x=\frac{4}{3},y=\frac{4}{3}$
• C. $C$ always passes through origin
• D. Equation of the director circle of the curve is $x^2+y^2=\frac{16}{9}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $C$ will be a hyperbola with eccentricity of $\sqrt{2}$
B $C$ will be a hyperbola whose asymptote are $x=\frac{4}{3},y=\frac{4}{3}$
C $C$ always passes through origin

Intersection point of $\frac{x}{4}+\frac{y}{8}=1$ and $\frac{x}{8}+\frac{y}{4}=1$ is $(\frac{8}{3},\frac{8}{3})$
Let the coordinate of the mid point of $AB$ is $(h,k)$ then
$A(2h,0)$ and $B(0,2k)$
$\because A,B,(\frac{8}{3},\frac{8}{3})$ are in straight line
$\therefore \Delta =0\Delta =\left|\begin{array}{ccc}2h& 0& 1\\ 0& 2k& 1\\ \frac{8}{3}& \frac{8}{3}& 1\end{array}\right|=04hk-\frac{16}{3}h-\frac{16}{3}k=03xy=4(x+y)$
Equation of the asymptotes are $x=\frac{4}{3},y=\frac{4}{3}$
Director circle of the hyperbola is point whose coordinates $(\frac{4}{3},\frac{4}{3})$