Find the equation of the tangent

Question:

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

Solution:

finding the slope of the tangent by differentiating the curve

$3 \frac{d y}{d x}+2 x=0$

$\frac{d y}{d x}=-\frac{2 x}{3}$

$\mathrm{m}($ tangent $)=-\frac{2 \mathrm{x}}{3}$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

now comparing the slope of a tangent with the given equation

$m($ tangent $)=4$

$-\frac{2 x}{3}=4$

$x=-6$

since this point lies on the curve, we can find y by substituting $x$

$6^{2}+3 y-3=0$

$y=-11$

therefore, the equation of the tangent is

$y+11=4(x+6)$

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