Find the equation of the tangent lines to the ellipse having a given angular coefficient Find the equations of the tangent lines to the ellipse $E : x^2/a^2 + y^2/b^2 − 1 = 0$ having a given angular coefficient $m ∈ R$.
 A: Here's a go at it:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Differentiate implicitly.
$$ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 $$
Assuming angular coefficient is the slope.
$$ \frac{dy}{dx} = m $$
$$ \frac{x}{a^2} + \frac{y}{b^2}m = 0 $$
$$ x  = - y \frac{a^2}{b^2}m $$
$$ x^2  = y^2 \frac{a^4}{b^4}m^2 $$
$$ y^2 \frac{a^2}{b^4}m^2 + \frac{y^2}{b^2} = 1 $$
$$ \frac{y^2}{b^2} \left[ \frac{a^2}{b^2}m^2 + 1 \right] = 1 $$
$$ y = \pm b \sqrt{ \frac{1}{ \left[ \frac{a^2}{b^2}m^2 + 1 \right] } } $$
You should be able to readily solve for the corresponding $x$ from here.
A: A line $\lambda x+\mu y+\tau = 0$ tangent to the ellipse satisfies the dual conic equation $$a^2\lambda^2+b^2\mu^2=\tau^2.\tag{*}$$ You’re given that the line has slope $m$, so its equation will be of the form $mx-y+\tau = 0$. Plug all of the known values into (*) and solve for $\tau$.  
Alternatively, if the line $y=mx+c$ is tangent to the ellipse, it has only one intersection with it. Substitute into the equation of the ellipse to get a quadratic equation in $x$. This must have a repeated root, so its discriminant must vanish. The latter constraint gives you a simple quadratic equation in $c$ to solve.
