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 

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

Answer 2

To determine which circle the points of intersection between the ellipse and the hyperbola lie on, follow these steps:

First, identify the given ellipse equation: \(4x^2 + 9y^2 = 36\). This can be rewritten in standard form by dividing the entire equation by 36:

\(\frac{x^2}{9} + \frac{y^2}{4} = 1\)

This standard form represents an ellipse with semi-major axis \(a = 3\) and semi-minor axis \(b = 2\). The focal distance \(c\) is found using \(c^2 = a^2 - b^2\):

\(c^2 = 3^2 - 2^2 = 9 - 4 = 5\). Thus, \(c = \sqrt{5}\).

Next, understand that the hyperbola is confocal with the ellipse. Hence, its equation has the same focal distance \(c = \sqrt{5}\) and takes the form:

\(\frac{x^2}{a'^2} - \frac{y^2}{b'^2} = 1\)

Given the transverse axis length (2) of the hyperbola, we know \(2a' = 2\), so \(a' = 1\). Using the formula \(c^2 = a'^2 + b'^2\), we solve for \(b'\):

\(5 = 1 + b'^2 \Rightarrow b'^2 = 4\).

Thus, the hyperbola's equation is:

\(\frac{x^2}{1} - \frac{y^2}{4} = 1\)

Now, determine the circle on which the intersection points lie. The circle equation is of the form \(x^2 + y^2 = r^2\).

The focal properties lead us to assess intersections using geometric symmetry, ultimately deducing that potential intersections reflect equidistant spacing from the center, based on \(c\). From this symmetry, deduce the circle:

Upon solving these conditions and using circle symmetry patterns in confocal conics, the correct circle where the ellipse and hyperbola intersect is:

\(x^2 + y^2 = 5\)

This matches the given option where the intersections are located.