Let the ellipse be:
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Set $P(a,t)$, the tangent line $PT_1$ is $y=k(x-a)+t$, $k=\tan A_1$, since it is tangent line, so put in ellipse equation, we have:
$$ b^2x^2+a^2(kx-ka+t)^2=a^2b^2$$
i.e.,
$$(b^2+a^2k^2)x^2+2a^2k(t-ka)x+a^2(t-ka)^2-a^2b^2=0$$
Solving for the point of tangency, the discriminant $\Delta$ must be zero:
$$\Delta$=$4a^4k^2(t-ka)^2-4(b^2+a^2k^2)[a^2(t-ka)^2-a^2b^2] $$
At $\Delta=0$, we get $ 2kta-t^2+b^2=0$, that is, $k=\dfrac{t^2-b^2}{2ta}$.
Let $k_1$ be $PF_1$'s slope, $k_1=\dfrac{t}{a+c}=\tan A_2$, the $ \angle T_1PF_1=|A_1-A_2|$, now we calculate $\tan(A_1-A_2)$:
$$\tan(A_1-A_2)=\dfrac{\tan A_1-\tan A_2}{1+\tan A_1 \tan A_2}=\dfrac{k-k_1}{1+k k_1}=\dfrac{\dfrac{t^2-b^2}{2ta}-\dfrac{t}{a+c}}{1+\dfrac{t^2-b^2}{2ta}\dfrac{t}{a+c}}=\dfrac{(t^2-b^2)(a+c)-2at^2}{2at^2+2act+t^3-tb^2}$$
Since $b^2=a^2-c^2$, we get:
$$ RHS=\dfrac{(t^2-a^2+c^2)(a+c)-2at^2}{t(2a^2+2ac+t^2-a^2+c^2)}=\dfrac{c-a}{t}$$
Let $k_2$ be line $PF_2$'s slope, then $k_2=\dfrac{t}{a-c}=\tan\angle PF_2T_2$. Since $\angle T_2PF_2=\dfrac{\pi}{2}-\angle PF_2T_2$, so $\tan \angle T_2PF_2=\dfrac{1}{\tan\angle PF_2T_2}=\dfrac{a-c}{t}=\tan(A_2-A_1)$. Since the angles are acute, we have:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$
The problem is that the given constraints determine $b$; you are not free to take $b=30$. As others have shown, the constraint on the tangent gives
$$ a = b\sqrt{\frac{200}{b-22}} $$
The constraint that $P$ be on the ellipse gives
$$ \frac{200^2}{a^2} + \frac{(b-22)^2}{b^2} = 1 $$
Use the first to eliminate $a$ from the second, and you'll get an equation for $b$, which is readily solved to give
$$ b = \frac{979}{14} \approx 69.9 $$
Best Answer
Translate the plane so that $D$ comes to the origin. The conic equation becomes
$$A(x-x_D)^2+B(x-x_D)(y-y_D)+C(y-y_D)^2+D(x-x_D)+E(y-y_D)+F=0$$
and you can compute the new coefficients.
Now assume that the equation of the tangent is
$$y=mx$$ and you get the condition
$$(cm^2+bm+a)x^2+(em+d)x+f=0.$$
We have tangency when this equation has a double root, i.e.
$$(em+d)^2-4f(cm^2+bm+a)=0.$$
This gives us the solutions
$$m=\frac{\pm\sqrt{(e^2-4cf) (4af-d^2)+(de-2 bf)^2}+2bf-de}{e^2-4cf}.$$