Question #220093

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.


1
Expert's answer
2021-07-26T15:58:31-0400

Given parabola -


y2=8xy^{2}=8x , the position vector of one end of an focal chord is 12i^+2j^\dfrac{1}{2}\hat{i}+2\hat{j}



We know that we let points on parabola as -


(at2,2at)=(12,2)(at^{2},2at)=(\dfrac{1}{2},2)


The value of a for given parabola is -


a=2a=2


So comparing the both sides of given equation , we get -


=at2=12=at^{2}=\dfrac{1}{2}


=t=12=t=-\dfrac{1}{2}


Now any other point on the parabola will be t1t_{1}


Now using the property of parabola , we get -


tt1=1tt_{1}=-1


So we get the value of t1=2t_{1}=2 .


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


(at12,2at1)=(8,8)(at_1^{2},2at_1)=(8,8)


so position vector at the other end will be -


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


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