The tangent at P to the curve y=x^2 has gradient 3. Find the equation of the normal at P.
The answer of this question is 12y=-4x+33
Please help me with this question, please
The slope of the tangent is given by $\displaystyle \frac{dy}{dx} = 2x$. If this is 3 then the x value is $\displaystyle x = \frac{3}{2}. $ The normal has slope $\displaystyle - \frac{1}{3}$. The equation for this through $\displaystyle \left( \frac{3}{2}, \frac{9}{4} \right)$ is
$\displaystyle
y - \frac{9}{4} = - \frac{1}{3} \left(x- \frac{3}{2} \right)
$
and if you simplify you'll get the answer you give.