Question

Line L₁ is parallel to the line L₂. Slope of L₁ is 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

60

Answer 2

The ellipse equation is $\frac{x^2}{25} + \frac{y^2}{9} = 1$, with $a^2 = 25$ and $b^2 = 9$. The general equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.

For the first pair of parallel lines with slope $m_1 = 9$: $y = 9x \pm \sqrt{25(9)^2 + 9} = 9x \pm \sqrt{2025 + 9} = 9x \pm \sqrt{2034}$. These lines can be written as $9x - y \pm \sqrt{2034} = 0$. Let $C_1 = \sqrt{2034}$ and $C_2 = -\sqrt{2034}$. Then $C_1 - C_2 = 2\sqrt{2034}$.

For the second pair of parallel lines with slope $m_2 = -\frac{1}{25}$: $y = -\frac{1}{25}x \pm \sqrt{25\left(-\frac{1}{25}\right)^2 + 9} = -\frac{1}{25}x \pm \sqrt{\frac{1}{25} + 9} = -\frac{1}{25}x \pm \sqrt{\frac{226}{25}} = -\frac{1}{25}x \pm \frac{\sqrt{226}}{5}$. Multiplying by 25, we get $25y = -x \pm 5\sqrt{226}$, which can be written as $x + 25y \mp 5\sqrt{226} = 0$. Let $C_3 = 5\sqrt{226}$ and $C_4 = -5\sqrt{226}$. Then $C_3 - C_4 = 10\sqrt{226}$.

The area of the parallelogram formed by lines $A_1x + B_1y + C_1 = 0$, $A_1x + B_1y + C_2 = 0$, $A_2x + B_2y + C_3 = 0$, $A_2x + B_2y + C_4 = 0$ is given by Area = $\frac{|(C_1 - C_2)(C_3 - C_4)|}{|A_1B_2 - A_2B_1|}$.

Here, $A_1 = 9$, $B_1 = -1$ for the first pair of lines. And $A_2 = 1$, $B_2 = 25$ for the second pair of lines. The denominator is $|A_1B_2 - A_2B_1| = |(9)(25) - (1)(-1)| = |225 + 1| = 226$. The numerator is $|(2\sqrt{2034})(10\sqrt{226})| = |20\sqrt{2034 \cdot 226}|$. $2034 \cdot 226 = (2 \cdot 3^2 \cdot 113) \cdot (2 \cdot 113) = 2^2 \cdot 3^2 \cdot 113^2 = (2 \cdot 3 \cdot 113)^2 = 678^2$. So, $\sqrt{2034 \cdot 226} = 678$. Area = $\frac{20 \cdot 678}{226} = \frac{20 \cdot (3 \cdot 226)}{226} = 20 \cdot 3 = 60$.