Question

The line y = x + 1 meets the ellipse \(\frac{x^2}{4}+\frac{y^2}{2}=1\) at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)² is equal to

• A. 20
• B. 12
• C. 11
• D. 8

Answer 1

The Correct Option is A

To solve the given problem, we need to find the radius \( r \) of the circle with the line segment \( PQ \) as its diameter. The line \( y = x + 1 \) intersects the ellipse \(\frac{x^2}{4} + \frac{y^2}{2} = 1\) at two points \( P \) and \( Q \). We will calculate the distance between these points, determine the radius of the circle, and subsequently compute \( (3r)^2 \).

First, substitute \( y = x + 1 \) into the ellipse equation:

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

Expanding and simplifying:

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

Multiply through by 4 to clear denominators:

\(x^2 + 2x^2 + 4x + 2 = 4\)

Combine and simplify terms:

\(3x^2 + 4x - 2 = 0\)

This is a quadratic equation in \( x \). Solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 4 \), and \( c = -2 \).

\(x = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-2)}}{2 \times 3}\) \(x = \frac{-4 \pm \sqrt{16 + 24}}{6}\) \(x = \frac{-4 \pm \sqrt{40}}{6}\) \(x = \frac{-4 \pm 2\sqrt{10}}{6}\) \(x = \frac{-2 \pm \sqrt{10}}{3}\)

Calculating for \( y \) using \( y = x + 1 \):

\(y_1 = \frac{-2 + \sqrt{10}}{3} + 1\) \(y_2 = \frac{-2 - \sqrt{10}}{3} + 1\)

Now, \( P: \left(\frac{-2 + \sqrt{10}}{3}, \frac{1 + \sqrt{10}}{3}\right) \) and \( Q: \left(\frac{-2 - \sqrt{10}}{3}, \frac{1 - \sqrt{10}}{3}\right) \).

The distance \( PQ \) (diameter) is calculated as:

\(PQ = \sqrt{\left(\frac{-2 + \sqrt{10}}{3} - \frac{-2 - \sqrt{10}}{3}\right)^2 + \left(\frac{1 + \sqrt{10}}{3} - \frac{1 - \sqrt{10}}{3}\right)^2}\) \(PQ = \sqrt{\left(\frac{2\sqrt{10}}{3}\right)^2 + \left(\frac{2\sqrt{10}}{3}\right)^2}\) \(PQ = \sqrt{\frac{40}{9} + \frac{40}{9}}\) \(PQ = \sqrt{\frac{80}{9}} = \frac{4\sqrt{5}}{3}\)

The radius \( r \) of the circle with \( PQ \) as diameter is half of \( PQ \):

\(r = \frac{2\sqrt{5}}{3}\)

Finally, compute \( (3r)^2 \):

\(3r = 2\sqrt{5}\) \( (3r)^2 = (2\sqrt{5})^2 = 20\)

Thus, \( (3r)^2 \) is equal to 20. Therefore, the correct answer is 20.