Question

Condition that two curves \( y^2 = 4ax \) and \( xy = c^2 \) cut orthogonally is:

• A. \( c^2 = 16a^2 \)
• B. \( c^2 = 32a^2 \)
• C. \( c^4 = 16a^4 \)
• D. \( c^4 = 32a^4 \)

Hint:

Use the product of slopes equal to -1 to find orthogonal intersection conditions.

Answer 1

The Correct Option is D

Differentiate \( y^2 = 4ax \) to get \( \frac{dy}{dx} = \frac{2a}{y} \).
Differentiate \( xy = c^2 \) implicitly to get \( \frac{dy}{dx} = -\frac{y}{x} \).
Apply the orthogonality condition: \[ \frac{2a}{y} \cdot \left(-\frac{y}{x}\right) = -\frac{2a}{x} = -1 \Rightarrow x = 2a \] Substitute \( x = 2a \) into \( xy = c^2 \) and \( y^2 = 4a(2a) = 8a^2 \Rightarrow y = \sqrt{8}a \).
Now \( c^2 = xy = 2a \cdot \sqrt{8}a = 2a \cdot 2\sqrt{2}a = 4\sqrt{2}a^2 \Rightarrow c^4 = 32a^4 \).