Answer to Question #223985 in Analytic Geometry for Bless

Question

If the position vector of one end of a chord through the focus of the parabola y²= 8x is 1/2i + 2j, find the position vector of the other end.

Answer 1

Let $x_0i+y_0j$ the position vector of the other end of a chord. Then $x_0=y_0^2/8$ .

The parameter of the parabola is

$$2p=8\Rightarrow \dfrac{p}{2}=2$$

The quadratic term in the equation is the y so the axis of the parabola coincides with the OX axis. Furthermore, the parabola lies on the positive side of the OX axis, since the coefficient accompanying the non-quadratic term (in this case the x) is 8 which is positive, so the focus lies at

$$F\left( \dfrac{p}{2},0\right)=F(2,0)$$

We have that:

$$\dfrac{2-0}{\frac12-2}=\dfrac{y_0-0}{y_0^2/8-2}$$

$$2\left( \dfrac{y_0^2}{8}-2\right) =-\dfrac32y_0$$

$$y_0^2+6y_0-16=0\Rightarrow y_0=-8,\quad y_0=2$$

The position vector of the other end is $y_0 = -8$ and $x_0 = y_0^2/8 = 8$

$$\boxed{8i-8j}$$

$$\dfrac{}{}$$