JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\mathrm{E}\) be an ellipse whose axes are parallel to the co-ordinates axes, having its center at \((3,-4)\), one focus at \((4,-4)\) and one vertex at \((5,-4) .\) If \(m x-y=4, m\,>\,0\) is a tangent to the ellipse \(\mathrm{E}\), then the value of \(5 \mathrm{~m}^{2}\) is equal to \(.....\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
and \(\mathrm{A}(5,-4)\) Hence, \(\mathrm{a}=2\, \,\mathrm{ae}=1\) \(\Rightarrow \mathrm{e}=\frac{1}{2}\) \(\Rightarrow \mathrm{b}^{2}=3\) So, \(E: \frac{(x-3)^{2}}{4}+\frac{(y+4)^{2}}{3}=1\) Intersecting with given tangent. \(\frac{x^{2}-6 x+9}{4}+\frac{m^{2} x^{2}}{3}=1\) Now,…
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