TS EAMCET · Maths · Hyperbola
If \(l\) is the maximum value of \(-3 x^2+4 x+1\) and \(m\) is the minimum value of \(3 x^2+4 x+1\), then the equation of the hyperbola having foci at \((l, 0),(7 \mathrm{~m}, 0)\) and eccentricity as 2 is
- A \(36 x^2-12 y^2=49\)
- B \(2 x^2-5 y^2=1\)
- C \(49 x^2-36 y^2=12\)
- D \(36 x^2-12 y^2=1\)
Answer & Solution
Correct Answer
(A) \(36 x^2-12 y^2=49\)
Step-by-step Solution
Detailed explanation
\(l = -3\left(\frac{4}{9}\right) + 4\left(\frac{2}{3}\right) + 1 = -\frac{4}{3} + \frac{8}{3} + 1 = \frac{4}{3} + 1 = \frac{7}{3}\) \(m = 3\left(\frac{4}{9}\right) + 4\left(-\frac{2}{3}\right) + 1 = \frac{4}{3} - \frac{8}{3} + 1 = -\frac{4}{3} + 1 = -\frac{1}{3}\) Foci:…
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