Question

Line L₁ is parallel to the line L₂. Slope of L₁ is $\frac{-1}{9}$. Also L₃ is parallel to L₄. Slope of L₄ is $\frac{-1}{25}$. All these lines touch the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. Find the area of the parallelogram formed by these lines.

Answer 1

The area of the parallelogram is $\frac{5\sqrt{42601}}{2}$.

Answer 2

The ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$, so $a^2=25$ and $b^2=9$. For lines with slope $m$, the tangent condition is $c^2 = a^2m^2 + b^2$. For $m_1 = -\frac{1}{9}$: $c_1^2 = 25\left(-\frac{1}{9}\right)^2 + 9 = \frac{25}{81} + 9 = \frac{754}{81}$, so $c_1 = \pm \frac{\sqrt{754}}{9}$. The lines are $x+9y \mp \sqrt{754} = 0$. For $m_2 = -\frac{1}{25}$: $c_2^2 = 25\left(-\frac{1}{25}\right)^2 + 9 = \frac{1}{25} + 9 = \frac{226}{25}$, so $c_2 = \pm \frac{\sqrt{226}}{5}$. The lines are $x+25y \mp 5\sqrt{226} = 0$. The area of the parallelogram formed by $a_1x+b_1y+d_1=0$, $a_1x+b_1y+d_2=0$, $a_2x+b_2y+d_3=0$, $a_2x+b_2y+d_4=0$ is $\frac{|(d_1-d_2)(d_3-d_4)|}{|a_1b_2-a_2b_1|}$. Area = $\frac{|(-\sqrt{754} - \sqrt{754})(-5\sqrt{226} - 5\sqrt{226})|}{|1 \cdot 25 - 1 \cdot 9|} = \frac{|(-2\sqrt{754})(-10\sqrt{226})|}{16} = \frac{20\sqrt{754 \cdot 226}}{16} = \frac{5\sqrt{754 \cdot 226}}{4}$. $754 \cdot 226 = (2 \cdot 377)(2 \cdot 113) = 4 \cdot 42601$. Area = $\frac{5\sqrt{4 \cdot 42601}}{4} = \frac{5 \cdot 2\sqrt{42601}}{4} = \frac{10\sqrt{42601}}{4} = \frac{5\sqrt{42601}}{2}$.