Find the equations of the tangent and the normal to the curve `4x^2+9y^2=36` at `(3costheta,\ 2sintheta)` at indicated points.

Explanation

`4 x^{2}+9 y^{2}=36`

On differentiating both sides w.r.t. x, we get

`8 x+18 y \frac{d y}{d x}=0`

`\text { or } 18 y \frac{d y}{d x}=-8 x`

`\text { or } \quad \frac{d y}{d x}=-\frac{8 x}{18 y}`

`\text { or } \quad \frac{d y}{d x}=\frac{-4 x}{9 y} \quad \ldots \text { (i) }`

But given that, tangent passes through the point `(3 \cos \theta, 2 \sin \theta).`


On putting ` x=3 \cos \theta, y=2 \sin \theta in Eq. (i),` we get

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