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10-2. Parabola, Ellipse, Hyperbola
hard
અહી ઉપવલય $E _1: \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1, a > b$ અને $E _2: \frac{ x ^2}{A^2}+\frac{ y ^2}{B^2}=1, A< B$ ની ઉત્કેન્દ્રિતા $\frac{1}{\sqrt{3}}$ સમાન છે. તેઓની નાભીલંભની લંબાઈનો ગુણાકાર $\frac{32}{\sqrt{3}}$ અને $E_1$ ની નાભીઓ વચ્ચેનું અંતર $4$ છે. જો $E_1$ અને $E_2$ એ બિંદુઓ $A, B, C$ અને $D$ આગળ છેદે છે તો ચતુષ્કોણ $A B C D$ નું ક્ષેત્રફળ મેળવો.
A$6 \sqrt{6}$
B$\frac{18 \sqrt{6}}{5}$
C$\frac{12 \sqrt{6}}{5}$
D$\frac{24 \sqrt{6}}{5}$
(JEE MAIN-2025)
Solution
$2 ae=4$
$2 a\left(\frac{1}{\sqrt{3}}\right)=4$
$\Rightarrow a=2 \sqrt{3}$
$\Rightarrow 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8$
$\text { Now } \frac{2 b^2}{a} \cdot \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 A^2}{B}=\frac{32}{\sqrt{3}}$
$\Rightarrow A^2=2 B$
$1-\frac{A^2}{B^2}=\frac{1}{3} \Rightarrow 1-\frac{2 B}{B^2}=\frac{1}{3} \Rightarrow B=3$
$\Rightarrow A^2=6$
$\frac{x^2}{12}+\frac{y^2}{8}=1 \ldots . .(1)$
$\frac{x^2}{6}+\frac{y^2}{9}=1 \ldots .(2)$
On solving $(1) \& (2)$ we get
$(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)$
The four points are vertices of rectangle and its area $= \frac{24 \sqrt{6}}{5}$
$2 a\left(\frac{1}{\sqrt{3}}\right)=4$
$\Rightarrow a=2 \sqrt{3}$
$\Rightarrow 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8$
$\text { Now } \frac{2 b^2}{a} \cdot \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 A^2}{B}=\frac{32}{\sqrt{3}}$
$\Rightarrow A^2=2 B$
$1-\frac{A^2}{B^2}=\frac{1}{3} \Rightarrow 1-\frac{2 B}{B^2}=\frac{1}{3} \Rightarrow B=3$
$\Rightarrow A^2=6$
$\frac{x^2}{12}+\frac{y^2}{8}=1 \ldots . .(1)$
$\frac{x^2}{6}+\frac{y^2}{9}=1 \ldots .(2)$
On solving $(1) \& (2)$ we get
$(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)$
The four points are vertices of rectangle and its area $= \frac{24 \sqrt{6}}{5}$
Standard 11
Mathematics
