Parametrics (1 Viewer)

nrlwinner · Feb 17, 2010

Hi. I got a question here is can't seem to solve.

Tangents at $P(x_1,y_1)$ and $Q(x_2,y_2)$ from the curve $x^2=4ay$ intersect at M $(\frac{2a(y_2-y_1)}{x_2-x_1},\frac{x_1y_2-x_2y_1}{x_2-x_1})$

Using $x_1^2=4ay_1 \and\ x_2^2=4ay_2$ , show M is $(\frac{x_1+x_2}{2},\frac{x_1x_2}{4a})$

I've got the x-cordinate of M, but I cant seem to get $\frac{x_1x_2}{4a}$

Drongoski · Feb 18, 2010

One way is to show point M lies on the tangents at P and at Q

$\100dpi T_{P}:\ \ \ xx_{1}\ =\ 2a(y+y_{1})\\ \\ T_{Q}:\ \ \ xx_{2}\ =\ 2a(y+y_{2})\\ \\ Now:\ \ \ \frac{x_{1}+x_{2}}{2}\cdot x_{1}\ =\ 2a(\frac{x_{1}x_{2}}{4a}\ +\ \frac{x_{1}^{2}}{4a})\\ \\ and:\ \ \ \frac{x_{1}+x_{2}}{2}\cdot x_{2}\ =\ 2a(\frac{x_{1}x_{2}}{4a}\ +\ \frac{x_{2}^{2}}{4a})\\ \\ i.e.\ \ the\ coords\ of\ M\ \ satisfy\ the\ eqns\ of\ both\ tangents.\\ \\ \therefore \ \ M \ is\ the\ point\ of\ intersection\ of\ the\ 2\ tangents.$

life92 · Feb 18, 2010

Okay so I can't type in Latex, and if I try to type it out here it'll be messy but..

As you can see in the answer you need to achieve, there is no 'y', so you just need to make the substitution y1 = x1^2 / 4a and y2 = x2^2 / 4a
Once you do this, simplify the expression and it should come out nicely.
I got the x and y values in 2-3 steps each.

If you want to see the solution, PM me.

Parametrics (1 Viewer)

nrlwinner

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Drongoski

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life92

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