# Math Help - Finding the eq of a normal to the parabola

1. ## Finding the eq of a normal to the parabola

Find the eq of the normal at a pt on the parabola $y^2 = 2x$ whose ordinate is 4

Solution

Eq of tangent to parabola at $(x_1,y_1)$ is $\Rightarrow yy_1 = 2a(x+x_1)$

Eq of parabola is $y^2 = 2x \therefore a = \frac{1}{2}$

\therefore the eq of a tangent to the parabola at $(x_1,4)$ is $\Rightarrow 4y = (x+x_1) \Rightarrow y = \frac{1}{4}(x+x_1)$

since the normal i perpendicular to the tangent then its slope would be -4

$\therefore$ eq of the normal to the parabola at a pt $(x_1, 4)$ is

$y = -4 (x+x_1)$

Am i correct !!!!!

2. Originally Posted by flintstone
Find the eq of the normal at a pt on the parabola $y^2 = 2x$ whose ordinate is 4

Solution

Eq of tangent to parabola at $(x_1,y_1)$ is $\Rightarrow yy_1 = 2a(x+x_1)$

Eq of parabola is $y^2 = 2x \therefore a = \frac{1}{2}$

\therefore the eq of a tangent to the parabola at $(x_1,4)$ is $\Rightarrow 4y = (x+x_1) \Rightarrow y = \frac{1}{4}(x+x_1)$

since the normal i perpendicular to the tangent then its slope would be -4

$\therefore$ eq of the normal to the parabola at a pt $(x_1, 4)$ is

$y = -4 (x+x_1)$

Am i correct !!!!!
No. and the fact that you have $x_1$ in your answer makes this obvious. the only variables should be x and y.

$y^2 = 2x$. When $y = 4$, $x = 8$, so the point the normal line passes through is (8,4). Now we need to find the slope.

Differentiating implicitly we see that $y' = \frac 1y$. Which means the slope when y = 4 is 1/4. which implies the slope of the normal line is -4.

I think you can take it from here.

(did you make an error? should this be in precalc or calc?)