JEE Main 2023 — Mathematics Coordinate Geometry

Given,

Equation of hyperbola,

$${H}_{n}:\frac{{x}^{2}}{1+n}-\frac{{y}^{2}}{3+n}=1,n\in \mathbb{N}$$Hn​:1+nx2​−3+ny2​=1,n∈N

Now using the eccentricity formula we get,

$${e}^{2}=1+\frac{3+n}{1+n}$$e2=1+1+n3+n​

$$\Rightarrow {e}^{2}=\frac{2n+4}{n+1}=\frac{2(n+2)}{n+1}$$⇒e2=n+12n+4​=n+12(n+2)​

Now check when $(n+1)=9,25,49,\ldots \ldots$(n+1)=9,25,49,…… is perfect square and then at the same time $2(n+2)$2(n+2) should also be perfect square,

So checking for $n=8\rightarrow {e}^{2}=\frac{20}{9}$n=8→e2=920​

$$n=24\rightarrow {e}^{2}=\frac{52}{25}$$n=24→e2=2552​

$$n=48\rightarrow {e}^{2}=\frac{100}{49}\Rightarrow e=\frac{10}{7}$$n=48→e2=49100​⇒e=710​

Hence, for $n=48$n=48 both n+1&2(n+2) are perfect square,

Now we know that,

Length of latusrectum is given by $l=\frac{2{b}^{2}}{a}=\frac{2(n+3)}{\sqrt{n+1}}=\frac{2\times 51}{7}$l=a2b2​=n+1​2(n+3)​=72×51​

Hence, $21l=\frac{42\times 51}{7}=306$21l=742×51​=306