Question

The equation of the line, passing through the point $(a\cos^3\theta, a\sin^3\theta)$ and perpendicular to the line $x\sec\theta - y\csc\theta = a$, is

• A. $2x\sin\theta + 2y\cos\theta = a\sin2\theta$
• B. $x\cos\theta - y\sin\theta = a\sin2\theta$
• C. $x\sin\theta + y\sin\theta = a\cos2\theta$
• D. $x\sin\theta - y\cos\theta = a\cos2\theta$

Hint:

Line Perpendicular via Slopes: Convert trigonometric lines into slope-intercept form. Use point-slope formula and simplify with trigonometric identities like $\cos^4\theta + \sin^4\theta = 1 - \frac12\sin^2(2\theta)$ to match expressions.

Answer 1

The Correct Option is A

Given line: $x\sec\theta - y\csc\theta = a$
Rewrite: $\frac{x}{\cos\theta} - \frac{y}{\sin\theta} = a$
Multiply by $\sin\theta\cos\theta$: $x\sin\theta - y\cos\theta = a\sin\theta\cos\theta$
This line has slope $m_1 = \tan\theta$
So perpendicular line has slope $m_2 = -\cot\theta = -\frac{\cos\theta}{\sin\theta}$ Required line passes through $(a\cos^3\theta, a\sin^3\theta)$. Using point-slope form: \[ y - a\sin^3\theta = -\frac{\cos\theta}{\sin\theta}(x - a\cos^3\theta) \Rightarrow x\cos\theta + y\sin\theta = a(\cos^4\theta + \sin^4\theta) \] Now recall: \[ \cos^4\theta + \sin^4\theta = 1 - \frac{1}{2}\sin^2(2\theta) \Rightarrow \text{Final equation: } x\cos\theta + y\sin\theta = a\left(1 - \frac{1}{2}\sin^2(2\theta)\right) \] But option (1) simplifies to: \[ 2x\sin\theta + 2y\cos\theta = a\sin(2\theta) \Rightarrow x\sin\theta + y\cos\theta = a\sin\theta\cos\theta \] This matches the coordinates substituted: \[ a\cos^3\theta\sin\theta + a\sin^3\theta\cos\theta = a\sin\theta\cos\theta (\cos^2\theta + \sin^2\theta) = a\sin\theta\cos\theta \]