Question

If the curves 2x = y² and 2xy = K intersect perpendicularly, then the value of K² is

• A. 4
• B. \(2\sqrt2\)
• C. 2
• D. 8

Answer 1

The Correct Option is D

If the curves \(2x = y^2\) and \(2xy = K\) intersect perpendicularly, then we need to find the value of \(K^2\).

First, find the derivatives of each curve.

For \(2x = y^2\), differentiate with respect to x:

\(2 = 2y \frac{dy}{dx}\)

\(\frac{dy}{dx} = \frac{1}{y}\). Let \(m_1 = \frac{1}{y}\).

For \(2xy = K\), differentiate with respect to x:

\(2x\frac{dy}{dx} + 2y = 0\)

\(2x\frac{dy}{dx} = -2y\)

\(\frac{dy}{dx} = -\frac{y}{x}\). Let \(m_2 = -\frac{y}{x}\).

For the curves to intersect perpendicularly, the product of their slopes must be -1:

\(m_1 \cdot m_2 = -1\)

\(\frac{1}{y} \cdot (-\frac{y}{x}) = -1\)

\(-\frac{1}{x} = -1\)

\(x = 1\)

Now, we substitute \(x = 1\) into the equation \(2x = y^2\):

\(2(1) = y^2\)

\(y^2 = 2\)

\(y = \pm \sqrt{2}\)

Next, we substitute \(x = 1\) and \(y = \pm \sqrt{2}\) into the equation \(2xy = K\):

\(2(1)(\pm \sqrt{2}) = K\)

\(K = \pm 2\sqrt{2}\)

Finally, we find \(K^2\):

\(K^2 = (\pm 2\sqrt{2})^2 = 4 \cdot 2 = 8\)

Therefore, the correct option is (D) 8.

Answer 2

Given curves are $ 2x = y^2 $ and $ 2xy = K $.

For $ 2x = y^2 $, differentiating with respect to $ x $, we have:

$$ 2 = 2y \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{y} = m_1 $$

For $ 2xy = K $, differentiating with respect to $ x $, we have:

$$ 2y + 2x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} = m_2 $$

Since the curves intersect perpendicularly, $ m_1 m_2 = -1 $.

$$ \left(\frac{1}{y}\right)\left(-\frac{y}{x}\right) = -1 \implies \frac{1}{x} = 1 \implies x = 1 $$

Since $ 2x = y^2 $, substituting $ x = 1 $:

$$ 2(1) = y^2 \implies y^2 = 2 \implies y = \pm \sqrt{2} $$

Since $ 2xy = K $, substituting $ x = 1 $ and $ y = \pm \sqrt{2} $:

$$ 2(1)(\pm \sqrt{2}) = K \implies K = \pm 2\sqrt{2} $$

Thus:

$$ K^2 = (2\sqrt{2})^2 = 4(2) = 8 $$