JEE Main 2022 — Mathematics Coordinate Geometry

Given,

Hyperbola:$\frac{{y}^{2}}{64}-\frac{{x}^{2}}{49}=1$64y2​−49x2​=1

And ellipse $E:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$E:a2x2​+b2y2​=1 passes through the vertices of the hyperbola $H:\frac{{x}^{2}}{49}-\frac{{y}^{2}}{64}=-1$H:49x2​−64y2​=−1, so vertices will be $V\equiv (0,\pm 8)$V≡(0,±8)

So ${b}^{2}=64$b2=64

Now eccentricity of hyperbola will be ${e}_{H}=\sqrt{1+\frac{{a}^{2}}{{b}^{2}}}=\sqrt{1+\frac{49}{64}}$eH​=1+b2a2​​=1+6449​​

And eccentricity of ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$a2x2​+b2y2​=1 will be

$${e}_{E}=\sqrt{1-\frac{{a}^{2}}{{b}^{2}}}=\sqrt{1-\frac{{a}^{2}}{64}}$$eE​=1−b2a2​​=1−64a2​​

And using $b=8$b=8

We get, ${e}_{H}\times {e}_{E}=\frac{1}{2}$eH​×eE​=21​ (given)

$$\Rightarrow \sqrt{\frac{1-{a}^{2}}{64}}\times \frac{\sqrt{113}}{8}=\frac{1}{2}$$⇒641−a2​​×8113​​=21​

$$\Rightarrow \sqrt{64-{a}^{2}}\times \sqrt{113}=32$$⇒64−a2​×113​=32

$$\Rightarrow (64-{a}^{2})=\frac{{32}^{2}}{113}$$⇒(64−a2)=113322​

$$\Rightarrow {a}^{2}=64-\frac{{32}^{2}}{113}$$⇒a2=64−113322​

Now length of latus rectum will be $l=\frac{2{a}^{2}}{b}=\frac{2}{8}(64-\frac{{32}^{2}}{113})=\frac{1552}{113}$l=b2a2​=82​(64−113322​)=1131552​

$$\Rightarrow 113l=1552$$⇒113l=1552