Find the number of normals to the parabola $y^2=8x$ through (2,1)
$$$$
I tried as follows:
Any normal to the parabola will be of the form $$y=mx-am^3-2am$$
Since the point (2,1) lies on the normal, it satisfies the equation of the normal. Thus $$2m^3+2m+1=0$$
The number of real roots of $m$ in the above equation would indicate the number of normals which satisfy the condition given in the question. However, I got stuck while trying to manually calculate the roots.
I would be grateful if somebody could please show me a more efficient method of solving the question. Many thanks in advance!
Best Answer
You have a cubic equation in $m$. You can find how many roots that equation has by looking at the discriminant
$$\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
for the equation $ax^3+bx^2+cx+d=0$. In your case, $a=2,\ b=0,\ c=2,\ d=1$. So your discriminant is
$$\Delta=0-0+0-4\cdot 2\cdot 2^3-17\cdot 2^2\cdot 1^2=-172$$
This is negative, so your equation has exactly one root. Your problem does not require actually finding the solution, so you are done (if, of course, your other work is correct).
For confirmation, you can look at the derivative of your equation:
$$\frac{dy}{dm}=6m^2+2$$
That derivative is always positive, so the function represented by $y$ is strictly increasing and there can be at most one zero. A cubic function must have at least one zero, so we know there is exactly one zero.