First slide

Tangents and normals

Question

The sum of intercepts on the axes of coordinates by any tangent to the curve _{$\sqrt{x} + \sqrt{y} = 2$} is

Difficult

A

2
B

4
C

8
D

_{$2 \sqrt{2}$}

Solution

For the curve _{$\sqrt{x} + \sqrt{y} = 2$} , the parametric coordinates are given by
_{$\sqrt{x} = 2 \cos^{2} θ$} and _{$\sqrt{y} = 2 \sin^{2} θ$} i.e., _{$x = 4 \cos^{4} θ$}

And _{$x = 4 \sin^{4} θ$}

_{$\frac{d y}{d x} = \frac{4 \times 4 \sin^{3} θ . \cos θ}{4 \times 4 \cos^{3} θ (- \sin θ)} = - \tan^{2} θ$}

_{$\Rightarrow E q .$} of tangent

_{$y - 4 \sin^{4} θ = - \tan^{2} θ (x - 4 \cos^{4} θ)$}

_{$∴ x - int e r c e p t = | 4 \cos^{4} θ + \frac{4 \sin^{4} θ}{\tan^{2} θ} | = 4 \cos^{2} θ a n d$}

_{$∴ y - int e r c e p t = | 4 \sin^{4} θ + \frac{4 \sin^{4} θ}{\tan^{2} θ} | = 4 \sin^{2} θ$}

Hence, the sum of intercept made on the axes of coordinates is _{$4 \cos^{2} θ + 4 \sin^{2} θ = 4$}

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