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: Understand the ellipse \(E_1\).
Given, \[ E_1=\frac{x^2}{9}+\frac{y^2}{4}=1 \] Comparing with the standard ellipse, \[ \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1 \] we get \[ \alpha=3,\quad \beta=2 \] So, the ellipse \(E_1\) touches the rectangle \(R\) at \[ x=\pm 3 \] and \[ y=\pm 2 \] Hence, the rectangle \(R\) has vertices \[ (\pm 3,\pm 2) \]

Step 2: Use the condition that \(E_2\) is circumscribed on \(R\).
Since \(E_2\) is circumscribed on \(R\), it passes through the vertices of \(R\).
Therefore, \(E_2\) passes through \[ (3,2) \] Given, \[ E_2=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] Substituting \((3,2)\), we get \[ \frac{9}{a^2}+\frac{4}{b^2}=1 \]

Step 3: Use the given point \((0,4)\).
Since \(E_2\) passes through \((0,4)\), we substitute \(x=0,\ y=4\) in \(E_2\).
\[ \frac{0^2}{a^2}+\frac{4^2}{b^2}=1 \] \[ \frac{16}{b^2}=1 \] Therefore, \[ b^2=16 \] So, \[ b=4 \]

Step 4: Find the value of \(a\).
Now substitute \[ b^2=16 \] in \[ \frac{9}{a^2}+\frac{4}{b^2}=1 \] So, \[ \frac{9}{a^2}+\frac{4}{16}=1 \] \[ \frac{9}{a^2}+\frac{1}{4}=1 \] \[ \frac{9}{a^2}=\frac{3}{4} \] Cross multiplying, \[ 3a^2=36 \] \[ a^2=12 \] Hence, \[ a=2\sqrt{3} \]

Step 5: Final conclusion.
Thus, \[ a=2\sqrt{3},\quad b=4 \] Therefore, \[ \boxed{a=2\sqrt{3},\ b=4} \]