Find the equation of the locus of the mid point of AB as m varies

Question

I am working through a pure maths book as a hobby. This question puzzles me.

The line y=mx intersects the curve $y=x^2-1$ at the points A and B. Find the equation of the locus of the mid point of AB as m varies.

I have said at intersection:

$mx = x^2-1 \implies x^2 - mx - 1=0$

Completing the square:

$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$

$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$

$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$

x-coordinates for points of intersection are

$x= \frac{-\sqrt(4+m^2) +m}{2}$ and $x= \frac{\sqrt(4+m^2)+m}{2}$

$\implies$ x-co-ordinate of P is mid-way between the two above points, namely $\frac{2m}{2} = m$

So $x=m, y = mx = m^2\implies y = x^2$

But my book says $y=2x^2$

Answer 1

You made a small mistake near the end: the midpoint's $x$-coordinate is not the sum of those of the intersection points, but half that sum, which is still $m/2$. Then the corresponding $y$-coordinate is $m^2/2$ and $y=2x^2$.

Answer 2

If $x_1$ and $x_2$ are x-coordinates of intersection points, x-coordinate of midpoint,

$x_m = \frac{x_1 + x_2}{2} = \frac{m}{2}$

$y = mx \implies y_m = 2 x_m^2$ so locus of midpoint is $y = 2 x^2$.

Your approach is correct but note that you can also get to it quickly using Vieta's formula for quadratic equation.

If $ax^2+bx+c = 0, x_1 + x_2 = - \frac{b}{a}, \ x_1 \ x_2 = \frac{c}{a}$

(where $x_1$ and $x_2$ are roots of the quadratic)

Here we have, $x^2 - mx - 1=0 \implies x_1 + x_2 = m \ $ so $x_m = \frac{m}{2}$.

Answer 3

The $x$-coordinate of the midpoint is the the sum of the other two divided by $2$, i.e. $$ x=\frac m2$$ And $$y=mx =\frac{m^2}{2} $$ and $$y=2x^2$$