The normal to the curve $x = 9(1 + \cos \theta), y = 9 \sin \theta$ at $\theta$ always passes through the fixed point}

Question

If the solution of \[ \left( 1 + 2e^\frac{x}{y} \right) dx + 2e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0 \] is \[ x + \lambda y e^\frac{x}{y} = c \quad \text{(where $c$ is an arbitrary constant), then $ \lambda $ is:} \]

Answer 1

Step 1: Understanding the Concept:
Identify the geometric locus of the parametric equations.
Normal to a circle always passes through its center.
Step 2: Key Formula or Approach:
$x - 9 = 9\cos\theta, y = 9\sin\theta$.
$(x-9)^2 + y^2 = 81(\cos^2\theta + \sin^2\theta) = 81$.
Step 3: Detailed Explanation:
This is a circle with center (9, 0) and radius 9.
Any normal to this circle will pass through the center.
Step 4: Final Answer:
The fixed point is (9, 0).

The normal to the curve $x = 9(1 + \cos \theta), y = 9 \sin \theta$ at $\theta$ always passes through the fixed point}

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The Correct Option is A

Solution and Explanation

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