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Q.
If the vertex of the conic represented by $25\left(x^{2}+y^{2}\right)=$ $(3 x-4 y+12)^{2}$ is $(a, b)$ then the value of $(b+a)$ is ______.
Conic Sections
Solution:
$25\left(x^{2}+y^{2}\right) =(3 x-4 y+12)^{2}$
$\Rightarrow \sqrt{x^{2}+y^{2}} =\left|\frac{3 x-4 y+12}{5}\right|$
which is equation of the parabola having focus at $A(0,0)$ and directrix
$L \equiv 3 x-4 y+12=0$ .... (i)
Now equation of axis of the parabola is $y=-\frac{4}{3} x$
or $4 x+3 y=0$ ..... (ii)
Solving (i) and (ii), we get point $B(-36 / 25,48 / 25)$
Now vertex is the mid point of $A B$, which is $(-18 / 25,24 / 25)$.
