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Question

If x2a2+y2b2=1, then d2ydx2=

A
b4a2y3
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B
b2ay2
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C
b3a2y3
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D
b3a2y2
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Solution

The correct option is A b4a2y3
Given, x2a2+y2b2=1
b2x2+a2y2=a2b2
On differentiating, we get
2a2yy1=2b2x
a2yy1=b2x
Again differentiating, we get
a2[yd2ydx2+(dydx)2]=b2
a2[yd2ydx2+b4x2y2a4]=b2

Thus we have : [yd2ydx2+b4x2y2a4]=b2a2

Substituting x2a2=1y2b2 in the above equation we have,

yd2ydx2+(b2y2)×b2a2y2=b2a2

d2ydx2=b2a2y(b2y2)×b2a2y3

Thus, d2ydx2=b2a2yb2×b2a2y3+b2a2y

d2ydx2=b4a2y3

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