Problem on finding the focus of parabola $4y^2+12x-20y+67=0$

Question

Find the focus of the parabola $4y^2+12x-20y+67=0$

$$$$

I tried proceeding as follows:

$$(2y-5)^2=-12x-42$$ $$(2y-5)^2=4(-3x-\frac{21}{2})$$

This is of the form $$Y^2=4aX$$ where $a=1, Y=2y-5, X=3x+\frac{21}{2}$.

Thus the focus is at $(a,0)$ with respect to the new coordinates $X$ and $Y$.

Hence, $$2y-5=0\Rightarrow y=\frac52$$ and $$-3x-\frac{21}{2}=1\Rightarrow x=\frac{-23}{6}$$

Unfortunately my value for $x$ is coming out to be incorrect. I would be grateful if somebody would kindly point out my mistake. Many thanks in advance!

Answer 1

Start by writing the original parabola as $$(y-\frac 52)^2=-3(x+\frac 72)$$comparing with $Y^2=4AX$ you should have $Y=y-\frac 52$, $4A=-3$ not $a=1$, so $X=x+\frac 72$.

Then the focus at $X=A$ becomes the focus at $x=-\frac 34-\frac 72$.

Similarly, $Y=0$ becomes $y-\frac 52=0$.