Question

If the ellipse \(4x^2 + 9y^2 = 36\) is confocal with a hyperbola whose length of the transverse axis is 2, then the points of intersection of the ellipse and hyperbola lie on the circle:

• A. \( x^2 + y^2 = 81 \)
• B. \( x^2 + y^2 = 16 \)
• C. \( x^2 + y^2 = 25 \)
• D. \( x^2 + y^2 = 5 \)

Hint:

For confocal conic sections, the foci remain the same. The points of intersection of an ellipse and its confocal hyperbola always lie on a circle centered at the origin with radius equal to the focal distance.

Answer 1

The Correct Option is D

We are given: - Ellipse: \( 4x^2 + 9y^2 = 36 \) - A confocal hyperbola whose transverse axis is 2. Step 1: Standard Form of the Ellipse Dividing the given ellipse equation by 36: \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] This is an ellipse with: - Semi-major axis \( a = 3 \) - Semi-minor axis \( b = 2 \) The focal distance \(c\) is calculated using the relation: \[ c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5} \] Step 2: Standard Form of the Hyperbola Since the given hyperbola is confocal with the ellipse, it must have the same focal distance \( c = \sqrt{5} \). The standard form of a hyperbola with this condition is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] We know the transverse axis is 2, so \( 2a = 2 \implies a = 1 \). Now using the identity for the focal distance in a hyperbola: \[ c = \sqrt{a^2 + b^2} \] Since \( c = \sqrt{5} \) and \( a = 1 \), \[ \sqrt{1 + b^2} = \sqrt{5} \] Squaring both sides: \[ 1 + b^2 = 5 \implies b^2 = 4 \] Thus, the hyperbola is: \[ \frac{x^2}{1} - \frac{y^2}{4} = 1 \] Step 3: Finding the Points of Intersection At points of intersection, add the two equations: \[ \frac{x^2}{9} + \frac{y^2}{4} + \frac{x^2}{1} - \frac{y^2}{4} = 1 + 1 \] Simplifying, \[ \frac{x^2}{9} + \frac{x^2}{1} = 2 \] Finding a common denominator: \[ \frac{x^2 + 9x^2}{9} = 2 \] \[ \frac{10x^2}{9} = 2 \] \[ x^2 = \frac{9}{5} \] From the ellipse equation: \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] Substituting \( x^2 = \frac{9}{5} \): \[ \frac{\frac{9}{5}}{9} + \frac{y^2}{4} = 1 \] \[ \frac{1}{5} + \frac{y^2}{4} = 1 \] \[ \frac{y^2}{4} = \frac{4}{5} \] \[ y^2 = \frac{16}{5} \] Step 4: Equation of the Circle Now compute \( x^2 + y^2 \): \[ x^2 + y^2 = \frac{9}{5} + \frac{16}{5} = \frac{25}{5} = 5 \] Step 5: Final Answer 

\[Correct Answer: (4) \ x^2 + y^2 = 5\]