Question

The length of the latus rectum and eccentricity of the Hyperbola \(9x^2-16y^2=144\) are

• A. \(\left(\frac{9}{4},\frac{5}{4}\right)\)
• B. \(\left(\frac{9}{2},\frac{5}{4}\right)\)
• C. \(\left(\frac{9}{2},\frac{5}{2}\right)\)
• D. \(\left(9,\frac{5}{2}\right)\)

Hint:

For \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), use \(e=\sqrt{1+\frac{b^2}{a^2}}\) and latus rectum \(=\frac{2b^2}{a}\).

Answer 1

The Correct Option is B

Concept: The standard form of hyperbola is: \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] For this hyperbola: \[ e=\sqrt{1+\frac{b^2}{a^2}} \] and length of latus rectum is: \[ \frac{2b^2}{a} \]

Step 1: Given equation: \[ 9x^2-16y^2=144 \] Divide both sides by \(144\): \[ \frac{9x^2}{144}-\frac{16y^2}{144}=1 \] \[ \frac{x^2}{16}-\frac{y^2}{9}=1 \]

Step 2: Compare with: \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] So: \[ a^2=16,\qquad b^2=9 \] \[ a=4,\qquad b=3 \]

Step 3: Find eccentricity. \[ e=\sqrt{1+\frac{b^2}{a^2}} \] \[ e=\sqrt{1+\frac{9}{16}} \] \[ e=\sqrt{\frac{25}{16}} \] \[ e=\frac{5}{4} \]

Step 4: Find length of latus rectum. \[ \frac{2b^2}{a} = \frac{2(9)}{4} \] \[ =\frac{18}{4} \] \[ =\frac{9}{2} \] Therefore, \[ \boxed{\left(\frac{9}{2},\frac{5}{4}\right)} \]