JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the ellipse \(3 x^2+\mathrm{py}^2=4\) pass through the centre \(C\) of the circle \(x^2+y^2-2 x-4 y-11=0\) of radius \(r\). Let \(f_1, f_2\) be the focal distances of the point C on the ellipse. Then \(6 f_1 f_2-r\) is equal to
- A 74
- B 68
- C 70
- D 78
Answer & Solution
Correct Answer
(C) 70
Step-by-step Solution
Detailed explanation
\(\mathrm{E}: \frac{\mathrm{x}^2}{4 / 3}+\frac{\mathrm{y}^2}{4 / \mathrm{P}}=1\) Centre of circle ( 1,2 ), radius \(\begin{aligned} & \mathrm{r}=\sqrt{1+4+11} \\ & \mathrm{r}=4 \end{aligned}\) \(\because\) E pass from centre \((1,2)\) \(\therefore \frac{3}{4}+\mathrm{P}=1\)…
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