Answer to Question #219591 in Analytic Geometry for Bless

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

Answer:-

Given parabola -

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

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 -

$=at^{2}=\dfrac{1}{2}$

$=t=-\dfrac{1}{2}$

Now any other point on the parabola will be $t_{1}$

Now using the property of parabola , we get -

$tt_{1}=-1$

{---- deducing the above property:

Coordinates of end point of focal chord are ($(at_1^{2},2at_1)$,$(at_2^{2},2at_2)$ at (a,0) and the focus is (a,0)

Three point are collinear , so

Slopes will be same, 

${2at_2-2at_1\over at_2^ 2- at_1^2}={2at_2-0\over at_2^2-a}$

${t_2-t_1\over t_2^2-t_1^2}={t_2\over t_2^2-t_1}$

solving we get

$t_1t_2=-1$ --------------}

So we get the value of $t_{1}=2$ .

Now , we know that other point on focal chord of parabola is -

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

so position vector at the other end will be -

$=8\hat{i}+8\hat{j}$ , which is required answer