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Differential Calculus> Q . SHOW THAT THE SUM OF THE INTERCEPTS ...

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Q . SHOW THAT THE SUM OF THE INTERCEPTS OF THE TANGENT TO THE CURVE -x1/2 + y1/2 = a1/2 ON THE COORDINATE AXES IS CONSTANT.

Bharat Makkar , 10 Years ago

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1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$\sqrt{x} + \sqrt{y} = \sqrt{a}$
Lets take the parametric coordinates of the curve,
$x = \frac{a}{4}cos^{2}\theta$
$\frac{dx}{d\theta } = \frac{a}{4}2cos\theta (-sin\theta)$
$y = \frac{a}{4}sin^{2}\theta$
$\frac{dy}{d\theta } = \frac{a}{4}2sin\theta cos\theta$
$\Rightarrow \frac{dy}{dx} = -1$
Equation of tangent:
$(y-\frac{a}{4}sin^{2}\theta ) = -(x-\frac{a}{4}cos^{2}\theta )$
$x + y = \frac{a}{4}$
Hence proved.

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