Find the equation of the tangent to the curve `x^2+3y-3=0` , which is parallel to the line `y=4x-5` .

Explanation

Suppose `(x_{1}, y_{1})` be the point of contact of tangent.

We can find the slope of the given line by differentiating the equation w.r.t `x` So, Slope of the line `=4`

Since,`(x_{1}, y_{1})` lies on the curve. Therefore,

`x_{1}^{2}+3 y_{1}-3=0 ldots` (1)

Now `^{2} x^{2}+3 y-3=0`

`Rightarrow 2 x+3 frac{d y}{d x}=0`

`Rightarrow frac{d y}{d x}=frac{-2 x}{3}`

Slope of tangent,`m=(frac{d y}{d x})_{(x_{1}, y_{1})}=frac{-2 x_{1}}{3}`

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