Question

The point \((p, q)\) is the point of intersection of a latus rectum and an asymptote of the hyperbola \(9x^2 - 16y^2 = 144\). If \(p>0\) and \(q>0\), then \(q = \ldots\)

• A. \(\frac{9}{4}\)
• B. \(\frac{7}{4}\)
• C. \(\frac{15}{4}\)
• D. \(\frac{13}{4}\)

Hint:

For a hyperbola, the asymptotes give the slope of the lines that intersect the latus rectum. The latus rectum is a line drawn through the focus that is perpendicular to the transverse axis.

Answer 1

The Correct Option is C

To solve this problem, we start by recalling the equation of the hyperbola: \[ 9x^2 - 16y^2 = 144. \] We can write it in the standard form by dividing through by 144: \[ \frac{x^2}{16} - \frac{y^2}{9} = 1. \] This represents a hyperbola with a horizontal transverse axis. The asymptotes of this hyperbola are given by: \[ y = \pm \frac{3}{4}x. \] The point \((p, q)\) is the intersection of the latus rectum and an asymptote. The equation for the latus rectum of a hyperbola is given by: \[ y = \pm \frac{b^2}{a}. \] Here, \(a^2 = 16\) and \(b^2 = 9\), so the latus rectum equation becomes: \[ y = \pm \frac{9}{4}. \] Thus, the value of \(q = \frac{15}{4}\) when \(p>0\) and \(q>0\). Hence, the correct answer is \(\boxed{\frac{15}{4}}\).