Question
If x2+y2=t−1t and x4+y4=t2+1t2,then x3ydydx equals
Answer: Option C
:
C
x2+y2=t−1t
On squaring both the sides
=x4+y4+2x2y2 = t2+1t2−2
Given that -
x4+y4=t2+1t2
∴x2y2=−1
⇒x2.2ydydx+y2.2x=0
⇒x3ydydx=−x2y2=1
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:
C
x2+y2=t−1t
On squaring both the sides
=x4+y4+2x2y2 = t2+1t2−2
Given that -
x4+y4=t2+1t2
∴x2y2=−1
⇒x2.2ydydx+y2.2x=0
⇒x3ydydx=−x2y2=1
Was this answer helpful ?
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