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Question : 145 of 200
Marks:
+1,
-0
Solution:
We have,
x=y2 . . . (i)
⇒‌2y‌‌=1⇒‌=‌‌‌ and ‌‌xy‌=a3 . . . (ii)
⇒x‌+y‌=0⇒‌=−‌On solving equations (i) and (ii), we get the point of intersection
(a2,a)m1=(‌)(a2,a)=‌and
‌‌m2=(‌)(a2,a)=‌=−‌Since, curves cut orthogonally then,
‌m1m2=−1‌⇒‌‌(‌)(−‌)=−1‌⇒‌‌a2=‌‌
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