If curves y²=4x and xy=c cut at right ... - CBSE Class 12 Math ...

CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 43 of 50

Marks: +1, -0

If curves

y^2=4x

and

xy=c

cut at right angles, then the value of

c

is

$4 \sqrt{2}$
8
$2 \sqrt{2}$
$-4 \sqrt{2}$

Solution: 👈: Video Solution

Explanation: Given curves,

y^2=4x

and

xy=c

cuts orthogonally.

Let they intersect at

(x_1, y_1)

.

Now,

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

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

\Rightarrow \left. \frac{dy}{dx} \right|_{(x_1,y_1)} = \frac{2}{y_1} \cdots (i)

\text{ and } xy=c

\therefore x \frac{dy}{dx}+y=0

\therefore \frac{dy}{dx} = \frac{-y}{x}

\Rightarrow \left. \frac{dy}{dx} \right|_{(x_1,y_1)} = -\frac{y_1}{x_1} \cdots (ii)

From eqs. (i) and (ii)

\frac{2}{y_1} \times \left(-\frac{y_1}{x_1}\right) = -1 \quad [\because m_1 m_2 = -1]

\Rightarrow x_1 = 2

\text{Put } x_1 = 2 \text{ in } y_1^2 = 4x_1 \text{, we get}

\Rightarrow y_1^2 = 4(2) = 8

y_1 = 2 \sqrt{2}

Now, put value of

x_1

and

y_1

in

x_1 y_1 = c

, we get

c = 2(2 \sqrt{2}) = 4 \sqrt{2}

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