# If $y=mx + c$ is a tangent to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $c^2=a^2m^2 + b^2$

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?

• you substitute $y=mx+c$ in ellipse equation and make the discriminant of the quadratic zero – Ekaveera Kumar Sharma Feb 22 '16 at 4:00

Substitute $y=mx+c$ into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have
For tangency, $\Delta=0$
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.
• Yes. $(x,mx+c)$ would represent a general point on the line. Get a quadratic in $x$, set the determinant $=0$. – Omanshu Thapliyal Feb 22 '16 at 4:10