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Question

The sum of the lengths of tangents and subtangent at a point of y=alog(x2a2),(a>0) is porportional to

A
|x|
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B
|y|
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C
|xy|
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D
|x/y|
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Solution

The correct option is C |xy|
y=alog(x2a2).a>0.
dydx=2axx2a2
Length of tangent =y   1+1(dydx)2
=1+(x2a2)2(2ax)2
=|y(x2+a2)||2ax|
Length of subtangent =ydxdy=∣ ∣ ∣ ∣y×1dydx∣ ∣ ∣ ∣
=y(x2a2)|2ax|
Sum =|x2y||2ax|=|xy|2a
Proportional to |xy|

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