Question

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

Answer 1

Given parabola $y^{2}=8x$ , the position vector of one end of an focal chord is $\dfrac{1}{2}\vec{i}+2\vec{j}.$

We know that we let points on parabola as

$$(at^{2},2at)=(\dfrac{1}{2},2)$$

The value of $a$ for given parabola is $a=2$.

So comparing the both sides of given equation , we get

$$at^{2}=\dfrac{1}{2}, 2at=2=>t=\dfrac{1}{2}$$

Any other point on the parabola will be $t_{1}.$

Suppose the coordinates of the extremity P of the focal chord are $(at^2, 2a),$ the coordinates of the other extremity Q of the focal chord are $(at_1^2, 2at_1),$ and the coordinates of the focus $F$ are $(a, 0).$

Then, PF and FQ,have the same slopes

$$\dfrac{2at-0}{at^2-a}=\dfrac{2at_1-0}{at_1^2-a}$$

$$tt_1^2-t=t_1t^2-t_1$$

$$tt_1(t_1-t)=t-t_1$$

We get

$$tt_{1}=-1$$

So we get the value of $t_{1}=-2$ .

Then other point on focal chord of parabola is

$$(at_1^{2},2at_1)=(8,-8)$$

So position vector at the other end will be

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