JEE Main 2024 — Mathematics Coordinate Geometry

Given conics are ${x}^{2}+{y}^{2}=4b\text{and}\frac{{x}^{2}}{16}+\frac{{y}^{2}}{{b}^{2}}=1$x2+y2=4band16x2​+b2y2​=1.

The intersection point of these conics lie on ${y}^{2}=3{x}^{2}$y2=3x2.

So, putting ${y}^{2}=3{x}^{2}$y2=3x2 in both the conics.

$$\Rightarrow {x}^{2}+3{x}^{2}=4b,\frac{{x}^{2}}{16}+\frac{3{x}^{2}}{{b}^{2}}=1$$⇒x2+3x2=4b,16x2​+b23x2​=1

$$\Rightarrow {x}^{2}=b,\frac{b}{16}+\frac{3b}{{b}^{2}}=1$$⇒x2=b,16b​+b23b​=1

$$\Rightarrow \frac{b}{16}+\frac{3}{b}=1$$⇒16b​+b3​=1

$$\Rightarrow {b}^{2}+48=16b$$⇒b2+48=16b

$$\Rightarrow {b}^{2}-16b+48=0$$⇒b2−16b+48=0

$$\Rightarrow {b}^{2}-12b-4b+48=0$$⇒b2−12b−4b+48=0

$$\Rightarrow (b-4)(b-12)$$⇒(b−4)(b−12)

$\Rightarrow b=4,12$⇒b=4,12 ( $b=4$b=4is rejected because curves coincide)

$$\Rightarrow b=12$$⇒b=12

$$\Rightarrow x=\pm \sqrt{12}$$⇒x=±12​

$$\Rightarrow 12+{y}^{2}=48$$⇒12+y2=48

$$\Rightarrow y=\pm 6$$⇒y=±6

Hence points of intersection are $(\pm \sqrt{12},\pm 6)$(±12​,±6).

So, length and breadth of rectangle are $2\sqrt{12}\text{and}12$212​and12.

So, the area of rectangle is given by, $A=2\sqrt{12}\times 12=48\sqrt{3}$A=212​×12=483​

$$\Rightarrow 3\sqrt{3}A=48\sqrt{3}\times 3\sqrt{3}$$⇒33​A=483​×33​

$$\Rightarrow 3\sqrt{3}A=432$$⇒33​A=432