JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the length of the latus rectum of the ellipse \(x^{2}+\) \(4 y^{2}+2 x+8 y-\lambda=0\) is \(4\) , and \(l\) is the length of its major axis, then \(\lambda+l\) is equal to\(......\)
- A \(72\)
- B \(73\)
- C \(74\)
- D \(75\)
Answer & Solution
Correct Answer
(D) \(75\)
Step-by-step Solution
Detailed explanation
\(\lambda+\ell=75\) \(x^{2}+4 y^{2}+2 x+8 y-\lambda=0\) \(\frac{(x+1)^{2}}{\lambda+5}+\frac{(y+1)^{2}}{\lambda+5}=1\) \(\because \frac{2 b^{2}}{a}=4\) \(\frac{2(\lambda+5)}{4}=4(\sqrt{\lambda+5})\) \(\lambda=59\) \(\lambda \neq-5\) \(l=2 a=2 \sqrt{\lambda+5}=2 \sqrt{65}=16\)…
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