Find the points on the curve `4x^2+9y^2=1` , where the tangents are perpendicular to the line `2y+x=0`

Answer

The required point are

`(frac{3}{2 sqrt{10}}, frac{1}{3 sqrt{10}})` and `(frac{-3}{2 sqrt{10}}, frac{-1}{3 sqrt{10}})`

Explanation

The equation of the given line is `y=-\frac{1}{2} x`

`\therefore` slope of this line `=-\frac{1}{2}`..(i)

Let the required point be `(x_{1}, y_{1})`

Now, `4 x^{2}+9 y^{2}=1 \Rightarrow 8 x+18 y \cdot \frac{d y}{d x}=0`


`\Rightarrow \frac{d y}{d x}=(\frac{-4 x}{9 y}) \Rightarrow(\frac{d y}{d
x})_{(x_{1}, y_{1})}=\frac{-4 x_{1}}{9 y_{1}}`


`\therefore` slope of the tangent at `(x_{1}, y_{1})=\frac{-4 x_{1}}{9 y_{1}} \ldots` (ii)


From (i) and (ii), we have


`\frac{-4 x_{1}}{9 y_{1}} \times(-\frac{1}{2})=-1` or `y_{1}=-\frac{2}{9} x_{1 . \text { (iii) }}`


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