Question #223985

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.


1
Expert's answer
2021-09-01T08:13:01-0400

Let x0i+y0jx_0i+y_0j the position vector of the other end of a chord. Then x0=y02/8x_0=y_0^2/8 .


The parameter of the parabola is


2p=8p2=22p=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(p2,0)=F(2,0)F\left( \dfrac{p}{2},0\right)=F(2,0)

We have that:


20122=y00y02/82\dfrac{2-0}{\frac12-2}=\dfrac{y_0-0}{y_0^2/8-2}

2(y0282)=32y02\left( \dfrac{y_0^2}{8}-2\right) =-\dfrac32y_0

y02+6y016=0y0=8,y0=2y_0^2+6y_0-16=0\Rightarrow y_0=-8,\quad y_0=2

The position vector of the other end is y0=8y_0 = -8 and x0=y02/8=8x_0 = y_0^2/8 = 8


8i8j\boxed{8i-8j}

\dfrac{}{}

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