Given an ellipse with foci $F_1, F_2$ and a point $P$. Let $T_1, T_2$ the points of tangency on the ellipse determined by the tangent lines through $P$. Show that $\widehat {T_1 P F_1} = \widehat {T_2 P F_2}$.
[Math] Equal angles formed by the tangent lines to an ellipse and the lines through the foci.
conic sectionsgeometry
Related Solutions
Let the foci be $F_1$ and $F_2$ and let a point of intersection be $P$. The tangent to the ellipse is the external angle bisector of $\angle F_1PF_2$, and the tangent to the hyperbola is the internal angle bisector.
Proving these statements without calculus is onerous. Here's a sketch of a partial proof for the statement about ellipses, to give the main ideas:
Let $F_1,F_2,P$ be as above. Let $b$ be the external bisector of $\angle F_1PF_2$; we want to show that $b$ is tangent to the ellipse. By the definition you've been given, that means we want to show that $b$ intersects the ellipse only at $P$. Consider first a point $Q$ on the other side of $b$ from the foci; I claim that $Q$ is not on the ellipse. To show this, draw $F_2Q$, meeting $b$ at $R$; also reflect $F_2$ in $b$ to obtain its image $F_2'$. Note that $F_1PF_2'$ are collinear because $b$ is the external angle bisector. Then $$ F_1Q+QF_2 = F_1Q + QR + RF_2 = F_1Q + QR + RF_2' > F_1F_2' = F_1P + PF_2' = F_1P + PF_2 $$ So $Q$ is not on the ellipse. I'm skipping some details here, notably why it has to be $>$ and not $\ge$ in the triangle inequality here. After you fill in that detail, and if you believe that the ellipse is a continuous curve, then we know that the ellipse doesn't cross $b$, so it is tangent to the ellipse. Then give a similar analysis for hyperbolas.
(We have not proved that conic sections have only one tangent at each point; I think I proved that once from your definition, but I only remember that it was unpleasant... and maybe my proof was wrong anyway.)
The nice way to prove these statements is with the notion of gradient from multivariable calculus: the direction of the gradient of the scalar field "distance from $F_i$" is obviously a unit vector pointing away from $F_i$; from this and the definition of an ellipse as a level curve of the sum of two such scalar fields it's clear that the normal to the ellipse bisects the desired angle. (On preview, along the lines of achille hui's answer.)
We may suppose that $$\text{the ellipse$\ :\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a\gt b\gt 0$}$$ $$\text{the circle$\ :\ x^2+y^2=a^2-b^2$}$$
As you wrote, we have $$2a=17\quad\Rightarrow \quad a=\frac{17}{2}$$
Since $$\frac{a^2-b^2-y^2}{a^2}+\frac{y^2}{b^2}=1\quad\Rightarrow\quad |y|=\frac{b^2}{\sqrt{a^2-b^2}}$$
we have $$30=\frac 12\times 2\sqrt{a^2-b^2}\times \frac{b^2}{\sqrt{a^2-b^2}}\quad\Rightarrow\quad b=\sqrt{30}.$$
Thus, the answer is $$2\sqrt{a^2-b^2}=2\sqrt{\left(\frac{17}{2}\right)^2-30}=\color{red}{13}.$$
Best Answer
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:
$\angle T_2PF_2=A_2-A_1=\angle T_1PF_1$