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