Question

Find the equation for the parabola with focus at (3, -3) and vertex at (3, 1).
a. x^2 + 6x +16y -7 = 0
b. x^2 + 6x + 16y - 25 = 0
c. x^2 - 6x - 16y + 25 = 0
d. x^2 - 6x + 16y -7 = 0

Answer 1

Vertex (3,1) and Focus (3,-3)
From these two we can deduce that the axis of the parabola is parallel to the x-axis.



Let,
\[X ^{2}=4aY\] be the eq. of the parabola.
\[X=(x-3) \]
\[Y=(y-1)\]
So, in real axis, the equation of the parabola is,
\[(x-3)^{2}=4a(y-1)\]
Now,
a=Distance between the vertex and the focus=\[\sqrt{(3-3)^{2} + (1-(-3))^{2}} = 4\]
Therefore the equation of the parabola is ,
\[(x-3)^{2}=16(y-1)\]
Solving which will give you,
\[x ^{2}-6x-16y+25=0\]

Therefore, (c) is the answer.