Question: Let $P:(x_0, y_0)$ be a first quadrant point on $\frac{x^2}{a^2} − \frac{y^2}{b^2} = 1, a>0, b>0$. Let $D$ be the point where the tangent at $P$ meets the line $x =a^2c$. $F_2$ is the focus $F_2: (c, 0), c>0, c=\sqrt{a^2+b^2}$. Show that $∠DF_2P$ is a right angle.
I believe the way to solve this problem is to confirm that the slope of $DP$ is the negative reciprocal of the slope of $DF_2$.But the points being $(x_0,y_0)$, $(a^2,y)$. and $(c,0)$ doesn't seem to get me there.
Hint: Try using parametric coordinates $( a\sec \theta ,b\tan \theta )$ for the point P.
Henceforth calculate the intersection of the tangent line $\frac{x\sec \theta }{a} -\frac{y\tan \theta }{b} =1$ with the line $x=a^{3} e$, assuming the focus of the hyperbola is (ae,0), where e is the eccentricity.
This intersection point turns out to be: $\left( a^{3} e,\frac{\left( a^{2}\sec \theta -1\right) b}{\tan \theta }\right)$.
So, we have P, D and F. Can you now confirm that the slope of DP is the negative reciprocal of the slope of DF?
