Solve by using equation of line in 2D

Question

Given that the line $x + 3y = 5$ is normal to the curve $y=x^2 + 5x + 6$ at a point $C$, i) find the coordinates of the point $C$ ii) find the equation of the tangent to the curve a the point $C$

Answer 1

Part 1) Put $y=\frac{5-x}{3}$ in the equation of curve and solve the quadratic equation for $x$ and then substitute for $y$

Now slope of your normal is $\frac{-5}{3}$, slope of your tangent will be $\frac{3}{5}$

Answer 2

$y=x^2+5x+6$

$\dfrac{dy}{dx}=2x+5$ let's say slope $(m_1)=2x+5$

From line equation slope $($say $m_2$$)$= $\dfrac{-1}{3}$

Now for lines to pe perpendicular $m_1 \times m_2 = -1$

From here can you do?