Write a Proof

The standard equation of hyperbola $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$has following known points
Center $C(h,k)$
Vertices $V(h\pm a,k)$
Foci $F(h\pm c,k)$
With Given Focal length equation as $c^2=a^2+b^2$

Actually the question is very simple if we use the fact that this equation $c^2=a^2+b^2$ is given to us.
We are given a little bit modified equation in our question as:
$$\frac{(x-h)^2}{b^2}-\frac{(y-k)^2}{a^2}=-1$$
The distances a,b and c are also given, so if we consider triangle $CV_2V_{c1}$ and apply pythagoras in it, we get
$$a^2+b^2=V_{c1}V_2^2$$
and $CF_2^2=c^2$
So, using the given relation $c^2=a^2+b^2$ we get
$$CF_2^2=V_{c1}V_2^2$$
or
$$\boxed{CF_2=V_{c1}V_2}$$
Hence Proved

But we can prove the relation $c^2=a^2+b^2$ which is the focal equation using the definition of hyperbola that
$$|d_2-d_1|=2a$$
So, if we take any two points (x,y) on the standard equation of hyperbola $$\frac{(x)^2}{a^2}-\frac{(y)^2}{b^2}=1$$ we get
$$|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}|=2a$$
$$\sqrt{(x+c)^2+y^2}=\sqrt{(x-c)^2+y^2}+2a$$ assuming $d_2>d_1$. Now square both the sides to get
$$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$
Expanding and cancelling terms we get
$$4xc-4a^2=4a\sqrt{(x-c)^2+y^2}$$
$$xc-a^2=a\sqrt{(x-c)^2+y^2}$$
Square both sides again to get
$$x^2c^2+a^4-2xca^2=a^2(x^2+y^2+c^2-2xc)$$
$$\frac{x^2c^2}{a^2}+a^2-2xc=x^2+y^2+c^2-2xc$$
$$\frac{x^2c^2}{a^2}+a^2=x^2+y^2+c^2$$
$$(\frac{c^2-a^2}{a^2})x^2-y^2=c^2-a^2$$
Dividing both sides with right hand side we get
$$\frac{(x)^2}{a^2}-\frac{(y)^2}{c^2-a^2}=1$$

We can write $c^2-a^2=b^2$ for some b, so we get
$$\frac{(x)^2}{a^2}-\frac{(y)^2}{b^2}=1$$ as standard equation which can be shifted to other point by translation of h and k units to get
$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$