Conic Sections Question 356
Question: The equation of a circle passing through the vertex and the extremities of the latus rectum of the parabola $ y^{2}=8x $ is
[MP PET 1998]
Options:
A) $ x^{2}+y^{2}+10x=0 $
B) $ x^{2}+y^{2}+10y=0 $
C) $ x^{2}+y^{2}-10x=0 $
D) $ x^{2}+y^{2}-5x=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
For parabola, $ y^{2}=8x $
therefore $ 4a=8 $
therefore $ a=2 $
vertex of $ y^{2}=8x,O\equiv (0,0) $
End points of latus-ractum $ L(a,2a);L’(a,-2a) $
therefore $ L(2,4);L’(2,-4) $
Circle passes through the point (0,0) (2,4) and (2,-4) Let equation of circle, $ x^{2}+y^{2}+2gx+2fy+c=0 $
After solving equation of circle, $ x^{2}+y^{2}-10x=0 $
Trick: option (c) satisfied by (2, 4) and (2,-4).
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