[Math] The normal line intersects a curve at two points. What is the other point

calculusderivativesimplicit-differentiation

The line that is normal to the curve $\displaystyle x^2 + xy – 2y^2 = 0 $ at $\displaystyle (4,4)$ intersects the curve at what other point?

I can not find an example of how to do this equation. Can someone help me out?

Best Answer

Given the curve $\mathcal{C}$ as

$$ x^2 + x y - 2 y^2 = 0. \tag 1 $$

Solve the curve $\mathcal{C}$

$$ x^2 + x y - 2 y^2 = 0.\\ \Downarrow\\ 4 x^2 + 4 x y - 8 y^2 = 0.\\ \Downarrow\\ \Big( 2 x + y \Big)^2 - 9 y^2 = 0.\\ \Downarrow\\ \Big( 2 x + y \Big)^2 = \Big( 9 y \Big)^2.\\ \Downarrow\\ 2 x + y = \pm 3 y.\\ \Downarrow\\ 2 x = \Big( \pm 3 - 1 \Big) y.\\ \Downarrow\\ y = \frac{2}{ \pm 3 - 1 } x.\\ \Downarrow\\ y = \frac{2}{2} x \vee y = \frac{2}{-4} x.\\ $$

So the curve $\mathcal{C}$ are two crossing lines, given by

$$ y = x \vee y = - \frac{1}{2} x. \tag{2} $$

The normal through point $(4,4)$

The normal through point $(4,4)$ is given by

$$ y = 8 - x. \tag 3 $$

Intersection

We need to find the intersection between $$ y = - \frac{1}{2} x $$ and $$ y = 8 - x. $$

So we get

$$ y = - \frac{1}{2} x = 8 - x.\tag 4 $$

Solving intersection

As $$ - \frac{1}{2} x = 8 - x, $$

we get $$ \frac{1}{2} x = 8, $$

so

$$ x = 16. $$

And we have $$ y = - \frac{1}{2} x, $$

so

$$ y = - 8. $$

Solution

The solution is given by

$$ \bbox[16px,border:2px solid #800000] { (x,y) = (16,-8). } \tag 5 $$