Question

The equation of the parabola whose vertex is at $(3,2)$ and whose focus is at $(5,2)$

• A. $y^2-8x+4y+12=0$
• B. $y^2-8x-4y+28=0$
• C. ​​​​​​$y^2+8x+4y-36=0$
• D. $y^2-8x+4y-28=0$

Answer 1

The correct option is B $y^2-8x-4y+28=0$

Let vertex $A(3,2)$ and focus is $S(5,2)$
Slope of $AS=\frac{2-2}{5-3}=0$ (which is parallel to $x$ axis)
Hence axis of parabola parallel to $x$ - axis.
The equation is of the form

$(y-k{)}^2=4a(x-h)$
or $(y-2{)}^2=4a(x-3)$
as $(h,k)$ is the vertex $(3,2)$
$a=$ distance between the focus and the vertex
$=\sqrt{(5-3{)}^2+(2-2{)}^2}=2$
Hence the required equation is
$(y-2{)}^2=8(x-3)\text{or}{y}^2-8x-4y+28=0$

Alternate Solution:
$\because y$ coordinates of the vertex and the focus both are same
$\therefore$ axis of the required parabola will be $y=2$
If we shift the origin at $(3,2)$, then the parabola will be the form of $Y^2=4aX$
and $a=\text{distance between focus and vertex}=2$
So required parabola will be
$(y-2{)}^2=8(x-3)$
$\Rightarrow y^2-8x-4y+28=0$