Question

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

• A. $y^2–8x–6y+25=0$
• B. $y^2–6x+8y–25=0$
• C. $x^2–6x–8y+25=0$
• D. $x^2+6x–8y–25=0$

Answer 1

The correct option is C

$x^2–6x–8y+25=0$

To fine the equation of parabola:

The given parabola is $y^2–12x–4y+4=0$

The given equation can be written as $y^2-4y+4=12x$or ${(y-2)}^2=4\left(3\right)x$

Thus here the vertex and foci are $(0,2)$ and $(3,2)$.

The vertex of the required parabola is $(3,2)=(h,k)$, focus is $(3,4)$and the axis of symmetry is $x=3$

$\Rightarrow a=\sqrt{{(3-3)}^2+{(2-4)}^2}=\sqrt{0+4}=2$

Thus the required equation of parabola is

$$\begin{gathered} \Rightarrow {(x-h)}^2=4a(y-k) \\ \Rightarrow {(x-3)}^2=4\times 2(y-2) \\ \Rightarrow x^2+9-6x=8y-16 \\ \Rightarrow x^2-6x-8y+25=0 \end{gathered}$$

Hence, the correct option is (C).