Question

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

• A. $y^{2}-8\,x-6\,y+25=0 $
• B. $y^2-6\,x+8\,y-25=0 $
• C. $ x^{2}-6\,x-8\,y+25=0 $
• D. $ x^{2}+6\,x-8\,y-25=0 $

Answer 1

Answer: (C)

Given equation can be rewritten as
$(y-2)^{2}=12 \,x$
Here, vertex and foci are $(0,2)$ and $(3,2)$.
$\therefore $ Vertex of the required parabola is $(3,2)$ and focus is $(3,4)$.
The axis of symmetry is $x=3$ and latusrectum $=4 \cdot 2=8$
Hence, required equation is
$(x-3)^{2}=8(y-2)$
or $x^{2}-6\,x-8\, y+25=0$