Question

If \( l_1 \) and \( l_2 \) are the lengths of the perpendiculars drawn from a point on the hyperbola \( 5x^2 - 4y^2 - 20 = 0 \) to its asymptotes, then \( \frac{l_1^2 l_2^2}{100} = \)

• A. \( \frac{20}{9} \)
• B. \( \frac{16}{81} \)
• C. \( \frac{4}{81} \)
• D. \( \frac{2}{9} \)

Hint:

When calculating the perpendicular distance from a point to an asymptote of a hyperbola, use the formula for the distance from a point to a line and simplify the expression to find the required value.

Answer 1

The Correct Option is C

Step 1: The equation of the hyperbola is given by \( 5x^2 - 4y^2 - 20 = 0 \). To rewrite it in standard form, divide both sides by 20: \[ \frac{x^2}{4} - \frac{y^2}{5} = 1. \] This represents a hyperbola with \(a^2 = 4\) and \(b^2 = 5\).
Step 2: The asymptotes of the hyperbola are given by the lines: \[ y = \pm \frac{\sqrt{5}}{2}x. \] These are the asymptotes of the hyperbola. 
Step 3: For a point \( (x_0, y_0) \) on the hyperbola, the perpendicular distances \( l_1 \) and \( l_2 \) from the point to the asymptotes can be calculated using the formula for the perpendicular distance from a point to a line. For the asymptotes, we have the equations \( y = \pm \frac{\sqrt{5}}{2}x \), so the formula for the perpendicular distance from \( (x_0, y_0) \) to these asymptotes is: \[ l_1 = \frac{|y_0 - \frac{\sqrt{5}}{2}x_0|}{\sqrt{1 + \left(\frac{\sqrt{5}}{2}\right)^2}} = \frac{|y_0 - \frac{\sqrt{5}}{2}x_0|}{\sqrt{1 + \frac{5}{4}}} = \frac{|y_0 - \frac{\sqrt{5}}{2}x_0|}{\frac{3}{2}} = \frac{2|y_0 - \frac{\sqrt{5}}{2}x_0|}{3}. \] Similarly, the other perpendicular distance \( l_2 \) is: \[ l_2 = \frac{|y_0 + \frac{\sqrt{5}}{2}x_0|}{\frac{3}{2}} = \frac{2|y_0 + \frac{\sqrt{5}}{2}x_0|}{3}. \] Step 4: Now, compute \( l_1^2 \) and \( l_2^2 \): \[ l_1^2 = \frac{4(y_0 - \frac{\sqrt{5}}{2}x_0)^2}{9}, \quad l_2^2 = \frac{4(y_0 + \frac{\sqrt{5}}{2}x_0)^2}{9}. \] Now, multiply these two expressions: \[ l_1^2 l_2^2 = \frac{16 \left( (y_0 - \frac{\sqrt{5}}{2}x_0)^2 (y_0 + \frac{\sqrt{5}}{2}x_0)^2 \right)}{81}. \] Using the difference of squares, we get: \[ l_1^2 l_2^2 = \frac{16 \left( y_0^2 - (\frac{\sqrt{5}}{2}x_0)^2 \right)^2}{81}. \] Step 5: Given that the equation of the hyperbola is \( \frac{x^2}{4} - \frac{y^2}{5} = 1 \), substitute the values for \( x_0 \) and \( y_0 \) that satisfy this equation. After solving, we find: \[ \frac{l_1^2 l_2^2}{100} = \frac{4}{81}. \]