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# Solution for Show that the Curves 4x = Y2 and 4xy = K Cut at Right Angles, If K2 = 512 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Show that the curves 4x = y2 and 4xy = k cut at right angles, if k2 = 512 ?

#### Solution

$\text { Given }:$

$4x = y^2 . . . \left( 1 \right)$

$4xy = k . . . \left( 2 \right)$

$\text { From (1) and (2), we get }$

$y^3 = k$

$\Rightarrow y = k^\frac{1}{3}$

$\text { From (1), we get }$

$4x = k^\frac{2}{3}$

$\Rightarrow x = \frac{k^\frac{2}{3}}{4}$

$\text { On differentiating (1) w.r.t.x, we get }$

$4 = 2y\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx} = \frac{2}{y}$

$\Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{4}, k^\frac{1}{3} \right) = \frac{2}{k^\frac{1}{3}} = 2 k^\frac{- 1}{3}$

$\text { On differentiating (2) w.r.t.x, we get }$

$4x\frac{dy}{dx} + 4y = 0$

$\Rightarrow \frac{dy}{dx} = \frac{- y}{x}$

$\Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{4}, k^\frac{1}{3} \right) = \frac{- k^\frac{1}{3}}{\left( \frac{k^\frac{2}{3}}{4} \right)} = - 4 k^\frac{- 1}{3}$

$\text { It is given that the curves intersect at right angles }.$

$\therefore m_1 \times m_2 = - 1$

$\Rightarrow 2 k^\frac{- 1}{3} \times - 4 k^\frac{- 1}{3} = - 1$

$\Rightarrow 8 k^\frac{- 2}{3} = 1$

$\Rightarrow k^\frac{- 2}{3} = \frac{1}{8}$

$\Rightarrow k^\frac{2}{3} = 8$

$\text { Cubing on both sides, we get }$

$k^2 = 512$

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Solution Show that the Curves 4x = Y2 and 4xy = K Cut at Right Angles, If K2 = 512 ? Concept: Tangents and Normals.
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