From the previous step, we have a = 2√2, so a² = 8. Since a² = α², we get α² = 8. Using ae = 3, we find the eccentricity e. Then, we use the relationship b² = a²(e² - 1) for a hyperbola. We know b² = 2α, so we substitute the values to verify consistency and find β². However, the problem statement uses β² in the final expression, but the hyperbola equation is given as (x²)/(α²) - (y²)/(2α) = 1. This implies b² = 2α. There seems to be a typo in the problem statement, where it should be 2α instead of 2α and β² in the final expression. Assuming the hyperbola equation is (x²)/(α²) - (y²)/(β²) = 1, then β² = 2α.

From the previous step, we have a = 2√2, so a² = 8. Since a² = α², we get α² = 8. Using ae = 3, we find the eccentricity e. Then, we use the relationship b² = a²(e² - 1) for a hyperbola. We know b² = 2α, so we substitute the values to verify consistency and find β². However, the problem statement uses β² in the final expression, but the hyperbola equation is given as (x²)/(α²) - (y²)/(2α) = 1. This implies b² = 2α. There seems to be a typo in the problem statement, where it should be 2α instead of 2α and β² in the final expression. Assuming the hyperbola equation is (x²)/(α²) - (y²)/(β²) = 1, then β² = 2α.

From the hyperbola equation, we have b² = 2α. Substituting the value of α = 2√2, we get β² = 4√2. Now we can substitute the values of α² and β² into the expression 2α² + 3β².

Let the circle $C$ touch the line $x - y + 1 = 0$ , have the centre on the positive $x$ -axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$ . Let $H$ be the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{2\alpha} = 1$ , whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$ . Then $2\alpha^2 + 3\beta^2$ is equal to

Answer: $16 + 12\sqrt{2}$

Step-by-step solution

Step 1: Find the center and radius of circle C

Let the center of the circle $C$ be $(h, 0)$ since it lies on the positive $x$ -axis. Let $r$ be its radius. The distance from the center $(h, 0)$ to the tangent line $x - y + 1 = 0$ is equal to the radius $r$ . We use the formula for the distance from a point to a line. Also, the chord length formula relates the radius, the distance from the center to the chord, and half the chord length. The distance from $(h, 0)$ to the chord line $-3x + 2y - 1 = 0$ is $d = \frac{|-3h - 1|}{\sqrt{(-3)^2 + 2^2}} = \frac{|3h+1|}{\sqrt{13}}$ . The chord length is given as $\frac{4}{\sqrt{13}}$ , so half the chord length is $\frac{2}{\sqrt{13}}$ . Using the Pythagorean theorem, $r^2 = d^2 + (\text{half chord length})^2$ .

Step 2: Solve for h and r

We equate the two expressions for $r^2$ and solve for $h$ . Since the center is on the positive $x$ -axis, we take the positive value of $h$ . Once $h$ is found, we can calculate the radius $r$ .

Step 3: Determine hyperbola parameters

The center of circle $C$ is $(3, 0)$ . The diameter of circle $C$ is $2r = 2(2\sqrt{2}) = 4\sqrt{2}$ . For the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{2\alpha} = 1$ , we have $a^2 = \alpha^2$ and $b^2 = 2\alpha$ . The length of the transverse axis is $2a$ , and one of the foci is $(ae, 0)$ . We are given that one focus of the hyperbola is the center of $C$ , so $ae = 3$ . We are also given that the length of the transverse axis is the diameter of $C$ , so $2a = 4\sqrt{2}$ .

Step 4: Calculate $\alpha$ and $\beta$

From the previous step, we have $a = 2\sqrt{2}$ , so $a^2 = 8$ . Since $a^2 = \alpha^2$ , we get $\alpha^2 = 8$ . Using $ae = 3$ , we find the eccentricity $e$ . Then, we use the relationship $b^2 = a^2(e^2 - 1)$ for a hyperbola. We know $b^2 = 2\alpha$ , so we substitute the values to verify consistency and find $\beta^2$ . However, the problem statement uses $\beta^2$ in the final expression, but the hyperbola equation is given as $\frac{x^2}{\alpha^2} - \frac{y^2}{2\alpha} = 1$ . This implies $b^2 = 2\alpha$ . There seems to be a typo in the problem statement, where it should be $2\alpha$ instead of $2\alpha$ and $\beta^2$ in the final expression. Assuming the hyperbola equation is $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ , then $\beta^2 = 2\alpha$ .

Step 5: Calculate $\beta^2$ and the final expression

From the hyperbola equation, we have $b^2 = 2\alpha$ . Substituting the value of $\alpha = 2\sqrt{2}$ , we get $\beta^2 = 4\sqrt{2}$ . Now we can substitute the values of $\alpha^2$ and $\beta^2$ into the expression $2\alpha^2 + 3\beta^2$ .

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