Question

Let \[ E_1=\frac{x^2}{9}+\frac{y^2}{4}=1 \] and \[ E_2=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] be two ellipses and \(R\) be a rectangle with sides parallel to the coordinate axes. Let \(E_1\) be inscribed ellipse in \(R\) and \(E_2\) be circumscribed ellipse on \(R\). If \(E_2\) passes through \((0,4)\), then

• A. \(a=4,\ b=2\sqrt{3}\)
• B. \(a=12,\ b=16\)
• C. \(a=16,\ b=16\)
• D. \(a=2\sqrt{3},\ b=4\)

Hint:

If an ellipse is inscribed in a rectangle with sides parallel to the coordinate axes, then the rectangle is formed by the tangent lines at the end points of the major and minor axes.

Answer 1

The Correct Option is D

Step 1: Locate the rectangle from E1.
\(E_1:\frac{x^2}{9}+\frac{y^2}{4}=1\) touches the sides of R at \(x=\pm3\) and \(y=\pm2\), so R has corners \((\pm3,\pm2)\).

Step 2: E2 must pass through these corners.
Since \(E_2\) circumscribes R, it passes through \((3,2)\):
\[\frac{9}{a^2}+\frac{4}{b^2}=1\]

Step 3: Use the extra given point.
\(E_2\) also passes through \((0,4)\):
\[\frac{16}{b^2}=1 \implies b^2=16 \implies b=4\]

Step 4: Solve for a.
\[\frac{9}{a^2}+\frac{4}{16}=1 \implies \frac{9}{a^2}=\frac{3}{4} \implies a^2=12 \implies a=2\sqrt{3}\]
\[\boxed{a=2\sqrt{3},\ b=4}\]