Question

The equation of straight line which passes through the point (a cos³θ, a sin³θ) and perpendicular to x secθ+y cosecθ = a is

• A. \(\frac{x}{a}+\frac{y}{b}=a\ \cos\theta\)
• B. x cosθ - y sinθ = a cos2θ
• C. x cosθ + y sinθ = a cos2θ
• D. x cosθ - y sinθ = -a cos2θ

Answer 1

The Correct Option is B

Given:
- The equation of the straight line is passing through the point \( (a \cos^3 \theta, a \sin^3 \theta) \) and is perpendicular to the line \( x \sec \theta + y \csc \theta = a \).
To find the equation of the line that satisfies these conditions, we use the fact that the general form of the equation of a line perpendicular to the given line can be derived from the condition of perpendicularity.
The given line has the equation:
\[ x \sec \theta + y \csc \theta = a \]
The slope of this line is the negative reciprocal of the slope of the perpendicular line. To derive the equation of the perpendicular line, we use the point of intersection and the general formula for a line in the form \( Ax + By + C = 0 \).
After solving, the correct equation of the perpendicular line passing through the given point is:
\[ x \cos \theta - y \sin \theta = a \cos 2\theta \]
Therefore, the correct answer is (B): \( x \cos \theta - y \sin \theta = a \cos 2\theta \)