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Question : 56 of 160
Marks:
+1,
-0
Solution:
‌e2=1−‌‌‌=‌‌b2=2a2Let
P≡(at2,2at)‌⇒‌‌‌+‌=1‌⇒‌‌t4+2t2=1Slope of tangent
PA to ellipse at
(at2,2at)‌‌+‌=1‌‌+‌‌=0‌⇒‌‌(‌)=‌−‌‌=‌=−tSlope of tangent
PB to parabola at
(at2,2at)‌y2=4ax‌2y‌=4a⇒‌=‌=‌Product of slopes is -1 .
⇒‌‌θ=90∘Coordinates of point
‌ is
‌(a‌cos(‌),bsin‌(‌))‌‌‌=(‌,‌×√2a)=(‌,√‌a)
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