Question

The line \(12x \cos \theta + 5y \sin \theta = 60\) is tangent to which of the following curves ?

• A. \(x^2 + y^2 = 60\)
• B. \(x^2 + y^2 = 169\)
• C. \(144x^2 + 25y^2 = 3600\)
• D. \(25x^2 + 12y^2 = 3600\)

Hint:

Convert the general equation of an ellipse to standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) to easily identify the semi-axes.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The condition for a line \(lx + my = n\) to be tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(a^2 l^2 + b^2 m^2 = n^2\).
Step 2: Detailed Explanation:
Given line: \(12 \cos \theta \cdot x + 5 \sin \theta \cdot y = 60\).
Here \(l = 12 \cos \theta\), \(m = 5 \sin \theta\), and \(n = 60\).
Let's analyze curve (C): \(144x^2 + 25y^2 = 3600 \implies \frac{x^2}{25} + \frac{y^2}{144} = 1\).
For this ellipse, \(a^2 = 25\) and \(b^2 = 144\).
Check the tangency condition:
\[ a^2 l^2 + b^2 m^2 = 25(12 \cos \theta)^2 + 144(5 \sin \theta)^2 \]
\[ = 25(144 \cos^2 \theta) + 144(25 \sin^2 \theta) = 3600 (\cos^2 \theta + \sin^2 \theta) = 3600 \]
Since \(n^2 = 60^2 = 3600\), the condition \(a^2 l^2 + b^2 m^2 = n^2\) is satisfied.
Therefore, the line is a tangent to curve (C).
Step 3: Final Answer:
The curve is \(144x^2 + 25y^2 = 3600\).