How to find the equation of a sleeping parabola?

Question

In one of the questions in my math textbook, I was given this equation and told that passes through (4,4) and (-4,8)
$$(y−k)^2=−4(x−h)$$

I understand this graph is a sideways parabola and the vertex is (h,k) and that I need to make two equations to solve for h and k I was able to create the equations: $$(4-k)^2=-4(4-h)$$ $$(8-k)^2=-4(-4-h)$$

Expanding gives: $16-8k+k^2=-16+4h$ and $64-16k+k^2=16+4h$ respectively.

In the first equation, I make h the subject: $$\frac{k^2}{4}-2k+8$$

But then where do I sub that in? If I sub into the first equation $(4-k)^2=-4(4-h)$, I will get $0=0$? I am very confused and I am not sure what I should do! I may have done the wrong step somewhere.There also seems to be no examples or solutions in the textbook, but they got an answer of $k=2$ and $h=5$

Any help with be greatly appreciated! Thank you!

Answer 1

Hint

You've got the right idea, but you simply back substituted your expression for $\ h\ $ into the wrong equation. You got a tautology when you did that because that equation was the same one from which you derived your expression for $\ h\ $ in the first place. Try substituting it instead into your second equation. That should give you an easily solvable equation for $\ k\ $.

Answer 2

Since you are solving the two equations simultaneously, you must substitute $h$ in the other equation and solve for $k$. $$64 -16k + k^{2} = 16 + 4(\frac{k^{2}}{4} - 2k + 8)$$ $$ \implies 8k = 16$$ $$ \implies k = 2$$

Now go on and substitute the value of $k$ into $\frac{k^{2}}{4} - 2k + 8$ to get $h$.

Answer 3

We have:

$$(4-k)^2=-4(4-h)\\ (8-k)^2=-4(-4-h)$$

Or

$$(4-k)^2=-4(4-h)\\ (8-k)^2=4(4+h)$$

By subtracting in pairs and making use of the identity $a^2-b^2=(a-b)(a+b)$, where $a=(8-k)^2, b=(4-k)^2$ we obtain:

$$(8-k)^2-(4-k)^2=4(4+h)+4(4-h)\Rightarrow\\ [(8-k)-(4-k)][(8-k)+(4-k)]=4(4+h+4=h)\Rightarrow \\ \\4(12-2k)=32\Rightarrow\\k=2$$

Substituting in either of the two equations gives $h=5$.