Question

Let $H_n: \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1, n \in N$. Let $k$ be the smallest even value of $n$ such that the eccentricity of $H_k$ is a rational number. If $l$ is the length of the latus rectum of $H_k$, then unit digit of $21l$ is equal to

Answer 1

6

Answer 2

The equation of the hyperbola is given by $H_n: \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$, where $n \in N$. Comparing this with the standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we have $a^2 = 1+n$ and $b^2 = 3+n$. The eccentricity $e$ of the hyperbola is given by $e^2 = 1 + \frac{b^2}{a^2}$. Substituting the values of $a^2$ and $b^2$, we get: $e^2 = 1 + \frac{3+n}{1+n} = \frac{(1+n) + (3+n)}{1+n} = \frac{4+2n}{1+n}$. We can rewrite this as $e^2 = \frac{2(n+1) + 2}{n+1} = 2 + \frac{2}{n+1}$. We are looking for the smallest even value of $n$ such that the eccentricity $e$ is a rational number. For $e$ to be rational, $e^2$ must be the square of a rational number. Let $e^2 = m^2$ for some rational number $m$. So, $m^2 = 2 + \frac{2}{n+1}$. Since $n \in N$, $n \ge 1$, so $n+1 \ge 2$. This implies $0 < \frac{2}{n+1} \le \frac{2}{2} = 1$. So, $2 < e^2 \le 2+1 = 3$. Thus, we are looking for a rational number $m$ such that $2 < m^2 \le 3$. Let $m = \frac{p}{q}$ where $p, q$ are coprime integers and $q \neq 0$. Then $2 < \frac{p^2}{q^2} \le 3$, which means $2q^2 < p^2 \le 3q^2$. From $m^2 = 2 + \frac{2}{n+1}$, we have $m^2 - 2 = \frac{2}{n+1}$. So, $n+1 = \frac{2}{m^2 - 2}$, which gives $n = \frac{2}{m^2 - 2} - 1 = \frac{2 - (m^2 - 2)}{m^2 - 2} = \frac{4 - m^2}{m^2 - 2}$. Substituting $m^2 = \frac{p^2}{q^2}$, we get: $n = \frac{4 - p^2/q^2}{p^2/q^2 - 2} = \frac{(4q^2 - p^2)/q^2}{(p^2 - 2q^2)/q^2} = \frac{4q^2 - p^2}{p^2 - 2q^2}$. We need $n$ to be an even positive integer. We can rewrite the expression for $n$: $n = \frac{4q^2 - p^2}{p^2 - 2q^2} = \frac{2(2q^2) - p^2}{p^2 - 2q^2}$. Let's try to express the numerator in terms of the denominator $p^2 - 2q^2$. $4q^2 - p^2 = - (p^2 - 4q^2) = - (p^2 - 2q^2 - 2q^2) = - (p^2 - 2q^2) + 2q^2$. So, $n = \frac{-(p^2 - 2q^2) + 2q^2}{p^2 - 2q^2} = -1 + \frac{2q^2}{p^2 - 2q^2}$. Since $n$ is a positive integer, $n \ge 1$. $-1 + \frac{2q^2}{p^2 - 2q^2} \ge 1 \implies \frac{2q^2}{p^2 - 2q^2} \ge 2$. Since $2q^2 > 0$, we must have $p^2 - 2q^2 > 0$. This is consistent with $m^2 > 2$. So, $2q^2 \ge 2(p^2 - 2q^2) \implies q^2 \ge p^2 - 2q^2 \implies 3q^2 \ge p^2$. This confirms our condition $2q^2 < p^2 \le 3q^2$. For $n = -1 + \frac{2q^2}{p^2 - 2q^2}$ to be an integer, $p^2 - 2q^2$ must be a divisor of $2q^2$. Let $d = p^2 - 2q^2$. Then $n = -1 + \frac{2q^2}{d}$. We need to find coprime integers $p, q$ such that $2q^2 < p^2 \le 3q^2$ and $n = -1 + \frac{2q^2}{p^2 - 2q^2}$ is the smallest even positive integer.

After testing values, we find that for $n=48$, $e^2 = 2 + \frac{2}{1+48} = 2 + \frac{2}{49} = \frac{98+2}{49} = \frac{100}{49} = \left(\frac{10}{7}\right)^2$. So for $n=48$, $e = \frac{10}{7}$, which is rational.

The smallest even value of $n$ for which $e$ is rational is $k=48$. For $H_k$, where $k=48$, we have $a^2 = 1+k = 1+48 = 49$ and $b^2 = 3+k = 3+48 = 51$. The length of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $l = \frac{2b^2}{a}$. Here $a = \sqrt{49} = 7$ and $b^2 = 51$. So, $l = \frac{2 \times 51}{7} = \frac{102}{7}$. We need to find the unit digit of $21l$. $21l = 21 \times \frac{102}{7} = 3 \times 102 = 306$. The unit digit of 306 is 6.