Question

Let E: $\frac{x^2}{16} + \frac{y^2}{\beta^2} = 1$ be the ellipse which cuts H orthogonally then $\beta$ is

• A. $4\sqrt{2}$
• B. $3\sqrt{2}$
• C. $2\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

$2\sqrt{2}$

Answer 2

The given hyperbola is H: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The eccentricity of H is $e = \sqrt{1 + \frac{b^2}{a^2}}$. The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. The eccentricity of the conjugate hyperbola is $e' = \sqrt{1 + \frac{a^2}{b^2}}$. Given that $e = e'$, so $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{a^2}{b^2}}$. This implies $\frac{b^2}{a^2} = \frac{a^2}{b^2}$, which means $a^4 = b^4$. Since $a, b > 0$, we have $a = b$. The hyperbola H is a rectangular hyperbola: $x^2 - y^2 = a^2$. The eccentricity is $e = \sqrt{1 + \frac{a^2}{a^2}} = \sqrt{2}$. The foci of the hyperbola are $(\pm ae, 0) = (\pm a\sqrt{2}, 0)$.

The locus of the foot of the perpendicular from the focus to any tangent of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the auxiliary circle $x^2 + y^2 = a^2$. Given that the foot of the perpendicular from a focus to one of the tangents is $(\frac{1}{2}, \frac{\sqrt{15}}{2})$. This point must lie on the auxiliary circle $x^2 + y^2 = a^2$. So, $(\frac{1}{2})^2 + (\frac{\sqrt{15}}{2})^2 = a^2$. $\frac{1}{4} + \frac{15}{4} = a^2$ which simplifies to $a^2 = 4$. Since $a=b$, we also have $b^2 = 4$. The hyperbola H is $x^2 - y^2 = 4$.

The circle C is $x^2 + y^2 = a^3 - 4b^2 + 12$. Substitute $a=2$ and $b=2$: $x^2 + y^2 = 2^3 - 4(2^2) + 12 = 8 - 4(4) + 12 = 8 - 16 + 12 = 4$. The circle C is $x^2 + y^2 = 4$.

Now, let's address the question: Let E: $\frac{x^2}{16} + \frac{y^2}{\beta^2} = 1$ be the ellipse which cuts H orthogonally then $\beta$ is. Hyperbola H: $x^2 - y^2 = 4$, which is $\frac{x^2}{4} - \frac{y^2}{4} = 1$. Here $A^2 = 4, B^2 = 4$. Ellipse E: $\frac{x^2}{16} + \frac{y^2}{\beta^2} = 1$. Here $a_e^2 = 16, b_e^2 = \beta^2$.

The condition for an ellipse $\frac{x^2}{a_e^2} + \frac{y^2}{b_e^2} = 1$ and a hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ to intersect orthogonally is $a_e^2 - b_e^2 = A^2 + B^2$.

Substituting the values: $16 - \beta^2 = 4 + 4$. $16 - \beta^2 = 8$. $\beta^2 = 16 - 8 = 8$. $\beta = \sqrt{8} = 2\sqrt{2}$.