Find the equation of the tangent lines to the ellipse having a given angular coefficient

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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 $m ∈ R$.

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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.

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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.