Question

If \(x = a \cos \theta\) and \(y = b \sin \theta\) then \(b^2x^2 + a^2y^2 =\)

• A. \(a^2b^2\)
• B. \(ab\)
• C. \(a^4b^4\)
• D. \(a^2+b^2\)

Hint:

When you see expressions like \(x = a \cos \theta\) and \(y = b \sin \theta\), think about how to isolate \(\cos \theta\) and \(\sin \theta\) (\(\cos \theta = x/a\), \(\sin \theta = y/b\)) and then use \(\cos^2\theta + \sin^2\theta = 1\). This gives \((x/a)^2 + (y/b)^2 = 1\), which is the equation of an ellipse.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves substituting parametric equations into an algebraic expression and simplifying it using a trigonometric identity. The goal is to eliminate the parameter \(\theta\).

Step 2: Key Formula or Approach:
We will substitute the given expressions for \(x\) and \(y\) into \(b^2x^2 + a^2y^2\) and use the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\).

Step 3: Detailed Explanation:
We are given:
\(x = a \cos \theta\)
\(y = b \sin \theta\)
The expression to evaluate is \(b^2x^2 + a^2y^2\).
Substitute the values of \(x\) and \(y\) into the expression:
\[ b^2x^2 + a^2y^2 = b^2(a \cos \theta)^2 + a^2(b \sin \theta)^2 \] Square the terms inside the parentheses:
\[ = b^2(a^2 \cos^2 \theta) + a^2(b^2 \sin^2 \theta) \] \[ = a^2b^2 \cos^2 \theta + a^2b^2 \sin^2 \theta \] Factor out the common term \(a^2b^2\):
\[ = a^2b^2 (\cos^2 \theta + \sin^2 \theta) \] Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ = a^2b^2 (1) \] \[ = a^2b^2 \]

Step 4: Final Answer:
The value of the expression is \(a^2b^2\). This matches option (A).