Condition for Tangency of an Ellipse

The question: for which $m,c$ does $y=mx+c$ touch the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, and where?

Write the line as $mx-y+c=0$ … (1). The tangent at a point $(x_1,y_1)$ on the ellipse is $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$, i.e. $\dfrac{x_1}{a^2}x+\dfrac{y_1}{b^2}y-1=0$ … (2).

If (1) is the tangent at $(x_1,y_1)$, then (1) and (2) represent the same line, so comparing coefficients of like terms:

Since $(x_1,y_1)$ lies on the ellipse: $\dfrac{x_1^2}{a^2}+\dfrac{y_1^2}{b^2}=1$. Substituting:

So the condition of tangency is $\boxed{c^2=a^2m^2+b^2}$, i.e. $c=\pm\sqrt{a^2m^2+b^2}$ — the line $y=mx\pm\sqrt{a^2m^2+b^2}$ is always tangent to the ellipse for any slope $m$, and the point of contact is

$\left(-\dfrac{a^2m}{c},\ \dfrac{b^2}{c}\right).$ …