# Finding the eq of a normal to the parabola

• Jan 4th 2010, 06:38 PM
flintstone
Finding the eq of a normal to the parabola
Find the eq of the normal at a pt on the parabola $\displaystyle y^2 = 2x$ whose ordinate is 4

Solution

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

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

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

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

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

$\displaystyle y = -4 (x+x_1)$

Am i correct !!!!!
• Jan 4th 2010, 06:46 PM
Jhevon
Quote:

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

Solution

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

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

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

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

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

$\displaystyle y = -4 (x+x_1)$

Am i correct !!!!!

No. and the fact that you have $\displaystyle x_1$ in your answer makes this obvious. the only variables should be x and y.

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

Differentiating implicitly we see that $\displaystyle y' = \frac 1y$. 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?)
• Jan 4th 2010, 06:51 PM
flintstone