Question

If the vertex of the conic represented by $25\left(x^2+y^2\right)=$ $(3x-4y+12{)}^2$ is $(a,b)$ then the value of $(b+a)$ is ______.

Answer 1

Answer: 0.24

$25\left(x^2+y^2\right)=(3x-4y+12{)}^2$
$\Rightarrow \sqrt{x^2+y^2}=\left|\frac{3x-4y+12}{5}\right|$
which is equation of the parabola having focus at $A(0,0)$ and directrix
$L\equiv 3x-4y+12=0$ .... (i)
Now equation of axis of the parabola is $y=-\frac{4}{3}x$
or $4x+3y=0$ ..... (ii)

Solving (i) and (ii), we get point $B(-36/25,48/25)$
Now vertex is the mid point of $AB$, which is $(-18/25,24/25)$.