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

$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$

Can you increase it :) I will show a proof ! Thanks

Ok!

Sorry it looks like I cant add to a question's total bounty if it has already been accepted to be answered.

thank you so much!!

Welcome, it would be great if you accept the answer if you are satisfied. Thanks

Oops! My bad first time using the site

thank you so much