AP EAMCET · Maths · Hyperbola
Let \(\mathrm{L}\left(\mathrm{x}_1, 4\right)\) 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 \(S\left(8, y_1\right)\) 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 \(4(\sqrt{17}+1)\)
Answer & Solution
Correct Answer
(B) \(4(\sqrt{17}-1)\)
Step-by-step Solution
Detailed explanation
We know that end of the latus rectum in lying in the first quadrant be (ae, \(\left.b^2 / a\right)\) and focus be \((c, 0)\) of hyperbola \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) \(\therefore\) According to question…
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