For the ellipse $b^2x^2+a^2y^2=a^2b^2$, the gradient of the tangent at $(a\cos\theta,\,b\sin\theta)$ is:

1. Differentiate $b^2x^2+a^2y^2=a^2b^2$ implicitly: $2b^2x+2a^2y\dfrac{dy}{dx}=0$. 2. Rearranging gives $\dfrac{dy}{dx}=-\dfrac{b^2x}{a^2y}$. 3. Substitute the point $x=a\cos\theta$, $y=b\sin\theta$. 4. $\dfrac{dy}{dx}=-\dfrac{b^2(a\cos\theta)}{a^2(b\sin\theta)}=-\dfrac{b\cos\theta}{a\sin\theta}$. 5. Differentiating the parametric equations separately gives the same result. The form $-\dfrac{a\sin\theta}{b\cos\theta}$ is the normal's gradient, the negative reciprocal. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 8: Coordinate Geometry II_