Suppose the equation of latus rectum is x=4 and the vertex is (2,3). I am confused wouldn't there be many parabola with this same vertex and latus rectum.If not how to find the equation?
The answer that is given is $$(y-3)^2=8(x-2)$$ in the textbook. I am not getting it. Thanks in advance.
Let $BE$ be the directrix
As the axis of the parabola is perpendicular to the latus rectum, the equation of the axis(OA) $y=3$
So, $c=a=3$
As the vertex O$(2,3)$ is the mid-point of $A,B, \frac{b+4}2=2\implies b=0$
So, the equation of the directrix(BE) will be $x=0$ as it is perpendicular to the axis.
If P$(h,k)$ be any point on the parabola,
the distance of $AP=\sqrt{(h-4)^2+(k-3)^2}$
the distance of P$(h,k)$ from the directrix $BE$ will be $|h|$
As the eccentricity of a parabola is $1$
So, $$\sqrt{(h-4)^2+(k-3)^2}=|h|$$
Squaring we get $(h-4)^2+(k-3)^2=h^2\implies (k-3)^2=h^2-(h-4)^2=8h-16=8(h-2)$
So, the equation of the parabola : $(y-3)^2=8(x-2)$