The straight line $y = m(x – a)$ will meet the parabola $y^2 = 4ax$ at two distinct real points for which values of $m$?

Question

The straight line $y = m(x – a)$ will meet the parabola $y^2 = 4ax$ at two distinct real points for which values of $m$?

The answer is given as $m \in \mathbb R - {\{0\}}$.

I tried to solve by using the method of discriminants as follows:

$\{m(x-a)\}^2=4ax$, and then putting $D>0$, I get $1+m^2>0$ which is true for all reals.

I can't understand why $m=0$ can't be true from the result I get using the discriminant method. Any ideas ?

Answer 1

Note that if we have a quadratic equation $ax^2+bx+c=0$, then the solution is given by: $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$

Hence, the solution implicitly assumes that $a \neq 0$, which is the assumption that the equation is exactly of second degree and no less.

Now, let us look at the equation $m^2(x-a)^2 = 4ax$. This expands to: $$ m^2x^2 - 2m^2xa + m^2a^2 = 4ax \implies m^2x^2 +x(-4a-2ma) + ma^2 = 0 $$

Hence, your discriminant evaluates to $m^2+1$, which is greater than zero, for all applicable $m$. We have already seen that $m \neq 0$ is not applicable for this situation, because in the general quadratic solution, $m$ appears in the denominator, hence if $m=0$ we will be dividing by zero, which is a problem.

At $m=0$, the equation is not quadratic, and there is only one solution. Hence, it follows that there are two solutions on $\mathbb R - \{ 0\}$.

Answer 2

The discriminant of a quadratic equation $ax^2+bx+c=0\ (\color{red}{a\not=0})$ is $D=b^2-4ac$.

In our case, we have $$m^2x^2+(-2am^2-4a)x+m^2a^2=0$$ For $m=0$, we have $-4ax=0$ which is not a quadratic equation.

Answer 3

$$ y^2=4ax $$ So we just need $a \ne 0$ to have a parabola. Otherwise there would be no possibility to have the intersection with a line be two distinct points.

For the intersection we got the system $$ y = m (x - a) \\ y^2 = 4ax $$ we note $m \ne 0$ is needed for two intersection points, otherwise we would only have $x=0, y=0$ as solution. For $m \ne 0$ we have $$ 4ax = m^2(x-a)^2 = m^2(x^2 -2 ax + a^2) \iff \\ 0 = x^2 - 2ax + a^2 - (4a/m^2) x \\ = x^2 - 2a(1 + 2/m^2) x + a^2 \\ = (x - (1+2/m^2)a)^2 - (1+2/m^2)^2a^2 + a^2 \\ = (x - (1+2/m^2)a)^2 - (4/m^2 + 4/m^4) a^2 \\ = (x - (1+2/m^2)a)^2 - (4/m^2)(1+1/m^2) a^2 \iff \\ (x - (1+2/m^2)a)^2 = (4/m^2)(1+1/m^2) a^2 \iff \\ x = (1+2/m^2)a \pm (2/m)\sqrt{1+1/m^2} a $$ This gives no new constraint for $m$. We just need $m \ne 0$.