Can we find the midpoint of a chord of this parabola?

Can we find the midpoint of a chord of this parabola?

Question

Given that $x_1$ and $x_2$ are the roots of $ax^2 + bx + c = 0$, state in terms of some or all of $a$, $b$, $c$: (i) the condition that $x_1 = x_2$, (ii) the value of $x_1 + x_2$.

1. Find the values of $m$ for which the line $y = mx$ is a tangent to the curve $y^2 = 3x - 1$.
2. The line $y = 2x$ meets the curve $3y = x^2 - 10$ at the points $A(x_1, y_1)$ and $B(x_2, y_2)$.
3. Obtain the quadratic equation whose roots are $x_1$ and $x_2$.
4. Without solving this equation, find the $x$ co-ordinate of the midpoint of $AB$.

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The red curve here is $y^2 = 3x-1$, while the blue line is $y = mx$.

Solution

We know that $a \neq 0$, since otherwise the equation would only have one root.

The discriminant of the quadratic equation $ax^2 + bx + c = 0$ is given by $b^2 - 4ac$.

There is one repeated root (that is, $x_1 = x_2$) if and only if $b^2 - 4ac = 0$, or equivalently $b^2 = 4ac$.

Since the roots of the quadratic equation $ax^2 + bx + c = 0$ are $x_1$ and $x_2$, we can rewrite the equation as $a(x - x_1)(x - x_2) = 0$.

Expanding this expression then yields $ax^2 - a(x_1 + x_2)x + ax_1x_2 = 0.$ Comparing these coefficients with the ones in the original equation yields $b = -a(x_1 + x_2)$ and hence $x_1 + x_2 = -\dfrac{b}{a}$.

By using the quadratic formula we get

$$\begin{align*} x_1 + x_2 &= \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a}\\ &= \frac{-b + \sqrt{b^2 - 4ac} - b - \sqrt{b^2 - 4ac}}{2a}\\ &= \frac{-2b}{2a}\\ &= -\frac{b}{a} \end{align*}$$

Alternative Solution

The line $y = mx$ is a tangent to the curve $y^2 = 3x - 1$ if the line touches the curve at exactly one point.

The intersection points are found by solving the equations of the line and the curve simultaneously.

Substituting $y=mx$ into the equation of the curve gives $m^2 x^2 = 3x - 1$, which is equivalent to the equation $m^2x^2 - 3x + 1 = 0$.

By part (i) above, we know that we have exactly one root if and only if $9 = 4m^2$, so $m = \pm \dfrac{3}{2}$.

As in part (a), the intersection points of a line with a curve are obtained by solving their equations simultaneously.

To find their $x$-coordinates, we substitute $y=3x$ into the equation of the curve to give $6x=x^2-10$, which rearranges to $x^2 - 6x - 10 = 0$.

We know that the $x$ coordinates of the points $A$ and $B$ are $x_1$ and $x_2$ respectively, so the midpoint of $AB$ has $x$-coordinate $\dfrac{x_1+x_2}{2}$.

By part (ii) above, we know that $x_1 + x_2 = 6$, since they are the roots of the equation $x^2 -6x - 10 = 0$. Thus the midpoint of $AB$ has $x$-coordinate $\dfrac{6}{2}=3$.