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Hyperbola question

2023 · 11 Apr · Shift 1 · Q42
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  5. /2023 · 11 Apr · Shift 1 · Q42

Hyperbola question

2023 · 11 Apr · Shift 1 · Q42

JEE MainMathematicsHyperbolaNumerical+4 / −1
Let Hn:x21+n−y23+n=1,n∈N\mathrm{H}_{\mathrm{n}}: \frac{x^{2}}{1+n}-\frac{y^{2}}{3+n}=1, n \in NHn​:1+nx2​−3+ny2​=1,n∈N. Let k\mathrm{k}k be the smallest even value of n\mathrm{n}n such that the eccentricity of Hk\mathrm{H}_{\mathrm{k}}Hk​ is a rational number. If lll is the length of the latus rectum of Hk\mathrm{H}_{\mathrm{k}}Hk​, then 21l21 l21l is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 306

  1. Write the hyperbola in standard form

Given Hn:x21+n−y23+n=1,H_n: \frac{x^2}{1+n}-\frac{y^2}{3+n}=1,Hn​:1+nx2​−3+ny2​=1, with n∈Nn\in \mathbb Nn∈N.

Compare with the standard hyperbola x2a2−y2b2=1.\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.a2x2​−b2y2​=1. So, a2=1+n,b2=3+n.a^2=1+n,\qquad b^2=3+n.a2=1+n,b2=3+n.

  1. Find the eccentricity

For the hyperbola x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,a2x2​−b2y2​=1, we have c2=a2+b2c^2=a^2+b^2c2=a2+b2 and hence eccentricity e=ca=1+b2a2.e=\frac{c}{a}=\sqrt{1+\frac{b^2}{a^2}}.e=ac​=1+a2b2​​.

Substituting a2=1+na^2=1+na2=1+n and b2=3+nb^2=3+nb2=3+n,

\sqrt{\frac{(n+1)+(n+3)}{n+1}}= \sqrt{\frac{2n+4}{n+1}}= \sqrt{\frac{2(n+2)}{n+1}}.$$ We need the **smallest even** value of $n$ such that $e$ is rational. 3. **Test even values of $n$** Let $n=2m$. Then $$e^2=\frac{2(n+2)}{n+1}=\frac{2(2m+2)}{2m+1}=\frac{4(m+1)}{2m+1}.$$ We need this to be a perfect square of a rational number. Check smallest even natural numbers: - For $n=2$: $$e^2=\frac{2(4)}{3}=\frac{8}{3},$$ so $e=\sqrt{\frac83}$, irrational. - For $n=4$: $$e^2=\frac{2(6)}{5}=\frac{12}{5},$$ irrational. - For $n=6$: $$e^2=\frac{2(8)}{7}=\frac{16}{7},$$ irrational. - For $n=8$: $$e^2=\frac{2(10)}{9}=\frac{20}{9},$$ irrational. - For $n=10$: $$e^2=\frac{2(12)}{11}=\frac{24}{11},$$ irrational. - For $n=12$: $$e^2=\frac{2(14)}{13}=\frac{28}{13},$$ irrational. - For $n=14$: $$e^2=\frac{2(16)}{15}=\frac{32}{15},$$ irrational. - For $n=16$: $$e^2=\frac{2(18)}{17}=\frac{36}{17},$$ irrational. - For $n=18$: $$e^2=\frac{2(20)}{19}=\frac{40}{19},$$ irrational. - For $n=20$: $$e^2=\frac{2(22)}{21}=\frac{44}{21},$$ irrational. - For $n=22$: $$e^2=\frac{2(24)}{23}=\frac{48}{23},$$ irrational. - For $n=24$: $$e^2=\frac{2(26)}{25}=\frac{52}{25},$$ irrational. - For $n=26$: $$e^2=\frac{2(28)}{27}=\frac{56}{27},$$ irrational. - For $n=28$: $$e^2=\frac{2(30)}{29}=\frac{60}{29},$$ irrational. - For $n=30$: $$e^2=\frac{2(32)}{31}=\frac{64}{31},$$ irrational. This direct checking is not efficient. So let us solve algebraically. 4. **Condition for rational eccentricity** We need $$\frac{2(n+2)}{n+1}$$ to be a square of a rational number. Let $$\frac{2(n+2)}{n+1}=\left(\frac{p}{q}\right)^2,$$ with integers $p,q$. Rewrite: $$2(n+2)=\left(\frac{p^2}{q^2}\right)(n+1).$$ A better substitution is to write $$e^2=2+\frac{2}{n+1}.$$ Indeed, $$\frac{2(n+2)}{n+1}=\frac{2n+4}{n+1}=2+\frac{2}{n+1}.$$ Let $n+1=m$. Then $m$ is odd (because $n$ is even), and $$e^2=2+\frac{2}{m}=\frac{2(m+1)}{m}.$$ For rational $e$, we require $$\frac{2(m+1)}{m}$$ to be a rational square. Now set $$\frac{2(m+1)}{m}=\frac{u^2}{v^2}.$$ Cross-multiplying, $$2v^2(m+1)=u^2 m.$$ This is a Pell-type condition. Instead of pursuing that, use the answer target through latus rectum and verify the minimal $n$. 5. **Length of latus rectum** For the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ the length of latus rectum is $$l=\frac{2b^2}{a}.$$ Here, $$a=\sqrt{n+1},\qquad b^2=n+3.$$ So $$l=\frac{2(n+3)}{\sqrt{n+1}}.$$ Hence $$21l=\frac{42(n+3)}{\sqrt{n+1}}.$$ If the stored answer is $306$, then $$\frac{42(n+3)}{\sqrt{n+1}}=306$$ which gives $$\frac{n+3}{\sqrt{n+1}}=\frac{306}{42}=\frac{51}{7}.$$ So $$7(n+3)=51\sqrt{n+1}.$$ Squaring, $$49(n+3)^2=2601(n+1).$$ Try $n+1=49$, i.e. $n=48$: $$e^2=\frac{2(50)}{49}=\frac{100}{49},$$ so $$e=\frac{10}{7},$$ which is rational. Also $n=48$ is even. Now check whether it is the **smallest** even such $n$. 6. **Show $n=48$ is the smallest even value** Let $n=2m$. Then $$e^2=\frac{4(m+1)}{2m+1}.$$ We want this to be a rational square. Since $\gcd(m+1,2m+1)=1$, $$\gcd(4(m+1),2m+1)=1$$ because $2m+1$ is odd. Thus the fraction $$\frac{4(m+1)}{2m+1}$$ is in lowest terms except for the factor $4$ in numerator, and for it to be a square of a rational number, numerator and denominator must each be perfect squares. So we need $$2m+1=s^2$$ for some odd integer $s$, and $$4(m+1)=t^2.$$ Since $4(m+1)$ is a square, $m+1$ must itself be a square. Let $$m+1=r^2.$$ Then $$2m+1=2(r^2-1)+1=2r^2-1=s^2.$$ So we need $$2r^2-s^2=1.$$ Testing smallest positive integers $r$: - $r=1\Rightarrow s^2=1$, giving $m=0$, hence $n=0$ (not in natural numbers if $\mathbb N=\{1,2,3,\dots\}$). - $r=2\Rightarrow s^2=7$ no. - $r=3\Rightarrow s^2=17$ no. - $r=4\Rightarrow s^2=31$ no. - $r=5\Rightarrow s^2=49=7^2$ yes. Thus $$m+1=r^2=25\Rightarrow m=24\Rightarrow n=48.$$ So the smallest even $n$ is indeed $$k=48.$$ 7. **Compute the latus rectum for $n=48$** Then $$a^2=49\Rightarrow a=7, \qquad b^2=51.$$ Hence $$l=\frac{2b^2}{a}=\frac{2\cdot 51}{7}=\frac{102}{7}.$$ Therefore $$21l=21\cdot \frac{102}{7}=3\cdot 102=306.$$ 8. **Final answer** $$\boxed{306}$$
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