Tangent with a Given Slope

Two tangents to x²/a² + y²/b² = 1 exist for each slope m. Point of contact: (∓a²m/√(a²m²+b²), ±b²/√(a²m²+b²)).

Substitute $y = mx + c$ into $x^2/a^2 + y^2/b^2 = 1$:

$$\frac{x^2}{a^2} + \frac{(mx+c)^2}{b^2} = 1$$ $$b^2x^2 + a^2(mx+c)^2 = a^2b^2$$ $$(b^2 + a^2m^2)x^2 + 2a^2mcx + a^2c^2 - a^2b^2 = 0$$

For tangency, discriminant $= 0$:

$$(2a^2mc)^2 - 4(b^2+a^2m^2)(a^2c^2 - a^2b^2) = 0$$ $$4a^4m^2c^2 - 4a^2(b^2+a^2m^2)(c^2-b^2) = 0$$ $$a^2m^2c^2 - (b^2c^2 - b^4 + a^2m^2c^2 - a^2m^2b^2) = 0$$ $$b^4 + a^2m^2b^2 - b^2c^2 = 0 \implies c^2 = a^2m^2 + b^2$$ $$c = \pm\sqrt{a^2m^2+b^2}$$

The two tangents: $y = mx \pm \sqrt{a^2m^2+b^2}$.

Point of contact: Substituting $c = \sqrt{a^2m^2+b^2}$ back and solving the quadratic (which now has a double root):

$$x = -\frac{a^2m}{\sqrt{a^2m^2+b^2}}, \quad y = \frac{b^2}{\sqrt{a^2m^2+b^2}}$$