JEE Main 2024 — Mathematics Coordinate Geometry

Given:

The equation of hyperbola is, ${x}^{2}-{y}^{2}{\mathrm{cosec}}^{2}\theta =5$x2−y2cosec2θ=5

$$\Rightarrow \frac{{x}^{2}}{5}-\frac{{y}^{2}}{5{\mathrm{sin}}^{2}\theta }=1$$⇒5x2​−5sin2θy2​=1

And the equation of ellipse is ${x}^{2}{\mathrm{cosec}}^{2}\theta +{y}^{2}=5$x2cosec2θ+y2=5

$\Rightarrow \frac{{x}^{2}}{5{\mathrm{sin}}^{2}\theta }+\frac{{y}^{2}}{5}=1$⇒5sin2θx2​+5y2​=1 ${\text{here}a<b\text{as}{\mathrm{sin}}^{2}\theta \leq 1}$herea<bassin2θ≤1

So, {e}_{H}=\sqrt{1+{\mathrm{sin}}^{2}\theta }&{e}_{E}=\sqrt{1-{\mathrm{sin}}^{2}\theta }

Also given, ${e}_{H}=\sqrt{7}{e}_{E}$eH​=7​eE​

$$\Rightarrow \sqrt{1+{\mathrm{sin}}^{2}\theta }=\sqrt{7}\sqrt{1-{\mathrm{sin}}^{2}\theta }$$⇒1+sin2θ​=7​1−sin2θ​

$$\Rightarrow 1+{\mathrm{sin}}^{2}\theta =7-7{\mathrm{sin}}^{2}\theta$$⇒1+sin2θ=7−7sin2θ

$$\Rightarrow 8{\mathrm{sin}}^{2}\theta =6$$⇒8sin2θ=6

$$\Rightarrow \mathrm{sin}\theta =\frac{\sqrt{3}}{2}$$⇒sinθ=23​​

$$\Rightarrow \theta =\frac{\pi }{3}$$⇒θ=3π​