1. ## Parametrics question

P is any variable point on the parabola x^2=-4y. THe tangent from P cuts the parabola x^2=4y at Q and R. Show that 3x^2=4y is the equation of the licus of the mid point of the chord RQ.

PS: How do you activate LateX editor??

2. ## Re: Parametrics question

use [ tex ] and [ /tex ] (remove the spaces inside the square brackets).

For example, [ tex ] x [ /tex ] gives $x$

The equation of the tangent at any point $(2t,\,-t^2)$ on the parabola $x^2 = -4y$ is given by:

\begin{aligned} y + t^2 &= -t(x-2t) \\ y&= t^2-tx\end{aligned}

Solving simultaneously the equations of the tangent and the parabola $x^2 = 4y$, we get

\begin{aligned} tx-t^2 &= \frac 14 x^2 \\x^2 + 4tx - 4t^2 &= 0 \end{aligned}

Treating this as a quadratic in $x$, we can solve it using the quadratic formula to obtain:

$Q\left(t(2\sqrt2 - 2),\,\frac 14 t^2(2\sqrt2 - 2)^2\right)$

$R\left(t(-2\sqrt2 - 2),\,\frac 14 t^2(-2\sqrt2 - 2)^2\right)$

Taking the midpoint and simplifying gives us the parametric equation

$\begin{cases} x &= -2t\\y &= 3t^2 \end{cases}$

and it's easy to show that the equivalent Cartesian equation is $3x^2 = 4y$.

(A nice trick you can use to find the $x$ midpoint in questions like these is to halve the sum of roots $\alpha + \beta = -b/a$ - this is most useful when you don't need the $y$ coordinates.)

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