Question

The equation of a normal to the curve \[ x = 4\sec\theta,\qquad y = 4\tan^2\theta \] at \( \theta = \frac{\pi}{4} \) is

• A. \( x + y\sqrt{2} = 7\sqrt{2} \)
• B. \( 2\sqrt{2}x + y = 8\sqrt{2} \)
• C. \( \sqrt{2}x + y = 7\sqrt{2} \)
• D. \( x + 2\sqrt{2}y = 12\sqrt{2} \)

Hint:

For parametric curves, always compute \( \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} \) before finding tangent or normal.

Answer 1

The Correct Option is D

Step 1: Find the point on the curve.
At \( \theta = \frac{\pi}{4} \): \[ x = 4\sec\frac{\pi}{4} = 4\sqrt{2},\qquad y = 4\tan^2\frac{\pi}{4} = 4 \]

Step 2: Compute the slope of the tangent.
\[ \frac{dx}{d\theta} = 4\sec\theta\tan\theta,\qquad \frac{dy}{d\theta} = 8\tan\theta\sec^2\theta \] \[ \Rightarrow \frac{dy}{dx} = \frac{8\tan\theta\sec^2\theta}{4\sec\theta\tan\theta} = 2\sec\theta \] At \( \theta=\frac{\pi}{4} \), \[ m_{\text{tan}} = 2\sqrt{2} \]

Step 3: Slope of the normal.
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tan}}} = -\frac{1}{2\sqrt{2}} \]

Step 4: Equation of the normal.
\[ y-4 = -\frac{1}{2\sqrt{2}}(x-4\sqrt{2}) \] \[ \Rightarrow x + 2\sqrt{2}y = 12\sqrt{2} \]

Step 5: Conclusion.
\[ \boxed{x + 2\sqrt{2}y = 12\sqrt{2}} \]