Question

The X and Y intercepts of the tangent to the hyperbola \(\frac{x^2}{20} - \frac{y^2}{5} = 1\) which is perpendicular to the line \(4x + 3y = 7\), are respectively

• A. \(\frac{-10}{3}, \frac{-5}{3}\)
• B. \(\frac{10}{3}, \frac{-5}{2}\)
• C. \(\frac{10}{3}, \frac{5}{2}\)
• D. \(\frac{10}{3}, \frac{5}{3}\)

Hint:

In conic questions, “intercepts” in MCQs often mean intercept lengths, which are taken positive.

Answer 1

The Correct Option is C

Concept:
The line \(4x+3y=7\) has slope \[ -\frac{4}{3} \] so any perpendicular tangent must have slope \[ \frac{3}{4} \] For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] a tangent with slope \(m\) is: \[ y=mx \pm \sqrt{a^2m^2-b^2} \] ip

Step 1: Write the tangent with slope \(\frac34\).
Here, \[ a^2=20,\qquad b^2=5,\qquad m=\frac34 \] So, \[ y=\frac34 x \pm \sqrt{20\left(\frac34\right)^2-5} \] \[ y=\frac34 x \pm \sqrt{20\cdot \frac{9}{16}-5} \] \[ y=\frac34 x \pm \sqrt{\frac{45}{4}-5} \] \[ y=\frac34 x \pm \sqrt{\frac{25}{4}} \] \[ y=\frac34 x \pm \frac52 \] ip

Step 2: Take the tangent with positive intercept lengths.
Choose \[ y=\frac34 x - \frac52 \] Its y-intercept is: \[ -\frac52 \] but the intercept length on Y-axis is \[ \frac52 \] For x-intercept, put \(y=0\): \[ 0=\frac34 x-\frac52 \] \[ \frac34 x=\frac52 \] \[ x=\frac{10}{3} \] ip

Step 3: Write the intercepts.
Thus the intercept lengths are: \[ \frac{10}{3},\ \frac{5}{2} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \frac{10}{3},\ \frac{5}{2}} \]