Download App

JEE Main 29-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2025 Paper - Shift 14 Marks
MCQ
Let the ellipse, $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$ and $\mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{\mathrm{~A}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~B}^{2}}=1, \mathrm{~A}<\mathrm{B}$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_{1}$ be 4. If $E_{1}$ and $E_{2}$ meet at $A, B, C$ and $D$, then the area of the quadrilateral ABCD equals:
  • A
    $6 \sqrt{6}$
  • B
    $\frac{18 \sqrt{6}}{5}$
  • C
    $\frac{12 \sqrt{6}}{5}$
  • D
    $\frac{24 \sqrt{6}}{5}$
✓

Answer

(D)
Sol. $\quad 2 \mathrm{ae}=4$
$2 \mathrm{a}\left(\frac{1}{\sqrt{3}}\right)=4$
$\Rightarrow \mathrm{a}=2 \sqrt{3}$
$\Rightarrow 1-\frac{\mathrm{b}^{2}}{12}=\frac{1}{3} \Rightarrow \mathrm{~b}^{2}=8$
Now $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}} \cdot \frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{32}{\sqrt{3}}$
$\Rightarrow A^{2}=2 B$
$1-\frac{\mathrm{A}^{2}}{\mathrm{~B}^{2}}=\frac{1}{3} \Rightarrow 1-\frac{2 \mathrm{~B}}{\mathrm{~B}^{2}}=\frac{1}{3} \Rightarrow \mathrm{~B}=3$
$\Rightarrow A^{2}=6$
$\frac{x^{2}}{12}+\frac{y^{2}}{8}=1$ ...(1)
$\frac{x^{2}}{6}+\frac{y^{2}}{9}=1$ ...(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 $=$ $\frac{24 \sqrt{6}}{5}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "SECTION - A [MATHS - MCQ]" from JEE Main 29-Jan-2025 Paper - Shift 1→JEE Main 29-Jan-2025 Paper - Shift 1 — all question types→JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $f:R \to R$, is a continuous function such that $\left| {f\left( x \right) - f\left( y \right)} \right| \geqslant \left| {{e^x} - {e^y}} \right|\forall x,y \in R$ then $f(x)$ is
→
Let $\int {{{\sec }^{ - 1}}\left[ {{{- \sin }^2}x} \right]dx}  = f\left( x \right) + C$ , (valid $x  \ne 0$ ) where $[k]$ denotes greatest integer less than or equal to $k$ and $f(0) = 0$ , then the value of ${\left( {f\left( {\frac{8}{{\pi x}}} \right)} \right)"}$ at $x = 2$ is (where $(')$ dash denotes the derivative)
→
$\mathop {{\rm{lim}}}\limits_{x \to 2} \left( {\frac{{\sqrt {1 - {\rm{cos}}\left\{ {2\left( {x - 2} \right)} \right\}} }}{{x - 2}}} \right)=$
→
If the arithmetic and geometric means of $a$ and $b$ be $A$ and $G$ respectively, then the value of $A - G$ will be
→
An ellipse having foci at $(3, 1)$ and $(1, 1) $ passes through the point $(1, 3),$ then its eccentricity is
→
If $\sin A,\sin B,\cos A$ are in $G.P.$, then roots of ${x^2} + 2x\cot B + 1 = 0$ are always
→
The value of $x$ for maximum value of $(\sqrt 3 \,\sin x + \cos x)$ is .....$^o$
→
If $y = \sin [\cos (\sin x)],$ then $dy/dx = $
→
Let a conic $\mathrm{C}$ pass through the point $(4,-2)$ and $\mathrm{P}(\mathrm{x}, \mathrm{y}), \mathrm{x} \geq 3$, be any point on $\mathrm{C}$. Let the slope of the line touching the conic $\mathrm{C}$ only at a single point $\mathrm{P}$ be half the slope of the line joining the points $P$ and $(3,-5)$. If the focal distance of the point $(7,1)$ on $C$ is $d$, then $12 \mathrm{~d}$ equals ...........
→
The coefficient of $x^r (0 \le r \le n - 1)$ in the expression :

$(x + 2)^{n-1} + (x + 2)^{n-2}. (x + 1) + (x + 2)^{n-3} . (x + 1)^2; + ...... + (x + 1)^{n-1}$ is :

→