Jee Main 2024 31 01 2024 Shift2 - Question14
Question 14
A parabola has vertex $(2,3)$, equation of directrix is $2 x-y=1$ and equation of ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, e=\frac{1}{\sqrt{2}}$ and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is
(1) $\frac{6564}{25}$
(2) $\frac{3288}{25}$
(3) $\frac{6272}{25}$
(4) $\frac{4352}{25}$
Show Answer
Answer: (4)
Solution:
Slope of axis $=\frac{1}{2}$
$y-3=\frac{1}{2}(x-2)$
$\Rightarrow 2 y-6=x-2$
$\Rightarrow 2 y-x-4=0$
$2 x+y-6=0$
$4 x+2 y-12=0$
$\alpha+1.6=4 \Rightarrow \alpha=2.4$
$\beta+2.8=6 \Rightarrow \beta=3.2$
Ellipse passes through $(2.4,3.2)$
$\Rightarrow \frac{\left(\frac{24}{10}\right)^{2}}{a^{2}}+\frac{\left(\frac{32}{10}\right)^{2}}{b^{2}}=1$
Also $1-\frac{a^{2}}{b^{2}}=\frac{1}{2}$
$\frac{a^{2}}{b^{2}}=\frac{1}{2}$
$\frac{144}{25} b^{2}+\frac{256}{25} a^{2}=a^{2} b^{2}$
$\frac{144}{25}+\frac{256}{25} \times \frac{1}{2}=a^{2}$
$\Rightarrow \frac{(128+144)}{25}=a^{2} \Rightarrow \frac{272}{25}=a^{2}$
$\Rightarrow b^{2}=\frac{2 \times 272}{25}$
Latus rectum $=\frac{2 b^{2}}{a}$
(Latus rectum) ${ }^{2}$
$$ =\frac{4 b^{4}}{a^{2}}=4\left(\frac{b^{2}}{a^{2}}\right) b^{2}=\frac{8 \times 272 \times 2}{25}=\frac{4352}{25} $$
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