Q.Find the equation of the tangent to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ at the point $(a\cos\theta,\,b\sin\theta)$.

Find the tangent slope by implicit differentiation, evaluate at $(a\cos\theta,b\sin\theta)$, form the point-slope line, then clear denominators and use $\sin^2\theta+\cos^2\theta=1$ to reach the clean standard form.

Differentiating $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$:

$\dfrac{2x}{a^2}+\dfrac{2y\,y'}{b^2}=0\Rightarrow y'=-\dfrac{b^2x}{a^2y}$

At $(a\cos\theta,b\sin\theta)$: $y'=-\dfrac{b^2(a\cos\theta)}{a^2(b\sin\theta)}=-\dfrac{b\cos\theta}{a\sin\theta}$

Tangent: $y-b\sin\theta=-\dfrac{b\cos\theta}{a\sin\theta}(x-a\cos\theta)$

Multiply by $a\sin\theta$: …