Question

Let \(L(x_1,4)\) be the end of the latus rectum of the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] lying in the first quadrant and let \(S(8,y_1)\) be the focus of the given hyperbola. Then the length of its transverse axis is:

• A. \(2(\sqrt{17}-1)\)
• B. \(4(\sqrt{17}-1)\)
• C. \(2(\sqrt{17}+1)\)
• D. \(4(\sqrt{17}+1)\)

Hint:

For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] remember: \[ c^2=a^2+b^2 \] and the endpoints of the latus rectum are \[ \left(\pm c,\pm \frac{b^2}{a}\right) \] These formulas are very important in coordinate geometry problems.

Answer 1

The Correct Option is B

Step 1: Recall key hyperbola relationships.
For $ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 $: foci at $ (\pm c, 0) $ where $ c^2 = a^2 + b^2 $. The latus rectum endpoints (first quadrant) are at $ \left(c, \dfrac{b^2}{a}\right) $.
Step 2: Extract $ c $ and $ b^2/a $ from given coordinates.
Focus $ S(8, y_1) $ lies on the $ x $-axis ($ y_1 = 0 $), so $ c = 8 $, giving $ a^2 + b^2 = 64 $. Latus rectum end $ L(x_1, 4) $ in first quadrant means $ x_1 = c = 8 $ and $ \dfrac{b^2}{a} = 4 $, so $ b^2 = 4a $.
Step 3: Substitute $ b^2 = 4a $ into $ a^2 + b^2 = 64 $.
\[ a^2 + 4a = 64 \Rightarrow a^2 + 4a - 64 = 0 \]
Step 4: Solve the quadratic for $ a $.
Using the quadratic formula: \[ a = \frac{-4 \pm \sqrt{16 + 256}}{2} = \frac{-4 \pm \sqrt{272}}{2} = \frac{-4 \pm 4\sqrt{17}}{2} = -2 \pm 2\sqrt{17} \] Since $ a > 0 $: $ a = 2(\sqrt{17} - 1) $.
Step 5: Compute the transverse axis length.
Transverse axis $ = 2a = 4(\sqrt{17} - 1) $.
Step 6: State the answer.
\[ \boxed{4(\sqrt{17} - 1)} \]