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  4. A parabola is drawn with its focus at (3,4) and vertex at the focus of the parabola y2-12 x-4 y+4=0. The equation of the parabola is:

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Q. A parabola is drawn with its focus at $(3,4)$ and vertex at the focus of the parabola $y^2-12 x-4 y+4=0$. The equation of the parabola is:

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Solution:

$y^2-12 x-4 y+4=0 $
$y^2-4 y=12 x-4 $
$(y-2)^2=12 x$
$Y^2=12 X$
focus : $X=A, Y=0$
image
$x=3, y=2$
$A(3,2)$
equ. of directrix is $y=0, PS = PM$
$\sqrt{(x-3)^2+(y-4)^2}=|y|$
by squaring, we will get $(x-3)^2+(y-4)^2=y^2$
$x^2-6 x-8 y+25=0$