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Question : 147 of 160
Marks:
+1,
-0
Solution:
Given curves are
y2=4x+4‌ and ‌y2=36(9−x) . . . (i)
On solving, we get the points
(8,6) and
(8,−6)On differentiating Eq. (i), we get
‌2y‌=4‌‌‌ and ‌‌‌2y‌=−36⇒‌‌‌=‌‌‌‌ and ‌‌‌‌=‌At point
(8,6),
m1=‌=‌=‌ and
m2=‌=‌=−3∴‌‌tan‌θ=‌=‌=∞⇒‌‌θ=‌And at point
(8,−6),
‌m1=‌=‌=−‌‌ and ‌m2=‌=‌=3‌∴‌‌tan‌θ=‌=‌=∞‌⇒‌‌θ=‌
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