If $y=mx + c$ is a tangent to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then show that $c^2=a^2m^2 + b^2$.
So for this question, first off I tried to differentiate it using implicit differentiation. I got $$\frac{dy}{dx}=-\frac{xb^2}{ya^2}$$
Afterward, I was unsure how to proceed. I tried subbing the above into $y=mx +c$, but I don't think that helps. How to I get $c^2=a^2m^2 + b^2$ out?
Substitute $y=mx+c$ into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have
\begin{align*} \frac{x^{2}}{a^{2}}+\frac{(mx+c)^{2}}{b^{2}} &=1 \\ \left( \frac{1}{a^{2}}+\frac{m^{2}}{b^{2}} \right)x^{2}+ \frac{2mc}{b^{2}} x+\frac{c^{2}}{b^{2}}-1 &=0 \end{align*}
For tangency, $\Delta=0$
\begin{align*} \left( \frac{2mc}{b^{2}} \right)^{2}- 4\left( \frac{1}{a^{2}}+\frac{m^{2}}{b^{2}} \right) \left( \frac{c^{2}}{b^{2}}-1 \right) &=0 \\ m^{2}a^{2}c^{2}-(b^{2}+a^{2}m^{2})(c^{2}-b^{2}) &= 0 \\ b^{4}-b^{2}c^{2}+a^{2}b^{2}m^{2} &=0 \\ a^{2}m^{2}+b^{2} &= c^{2} \end{align*}
Let a general point on the given line be represented by $P=(\alpha, m\alpha+c)$. Then this point's distance taken from the two foci ($(ae,0); (-ae,0)$) must sum to $2a$, where $e=(1-\frac{b^2}{a^2})^\frac{1}{2}$. $\therefore ((\alpha-ae)^2 + (\alpha m+c-0)^2))^\frac{1}{2} + ((\alpha m+ae)^2 + (\alpha+c-0)^2))^\frac{1}{2} = 2a$ But you don't actually solve need to solve this equation for $\alpha$. Get the quadratic equation in $\alpha$ and since the line $y=mx+c$ is a tangent, it touches the curve at a single point, hence the determinant of the equation in $\alpha$ should be zero. That should give you the result.
