Question

The mid point of the chord $4x-3y=5$ of the hyperbola $2x^2-3y^2-12$ is

• A. $0,-\frac{5}{3}$
• B. $2,1$
• C. $\frac{5}{4},0$
• D. $\frac{11}{4},2$

Answer 1

Answer: (B)

$4x-3y=5$
$2x^2-3y^2=12$
eliminate $y$ from line and put into the equation of hyperbola
$y=\left(\frac{4x-5}{3}\right)$
$2x^2-3{\left(\frac{4x-5}{3}\right)}^2=12$
$x_1,x_2$ are the roots of the above quadratic
Now,
we have to find mid point of chord, if co-ordinates of chord of contact are $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$
so co-ordinates of mid point is $(x_1+x_2)/2,(y_1+y_2)/2$
$x_1+x_2=$ sum of roots $=4$
so $x-$ coordinate of mid point is $\frac{4}{2}=2$
put this $x-$ coordinates in chord we get $y=1$
so mid point is $(2,1)$