Question

Let H: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is a hyperbola such that its eccentricity and eccentricity of its conjugate hyperbola are same. Let foot of the perpendicular drawn from focus of H to its one of the tangent is $(\frac{1}{2}, \frac{\sqrt{15}}{2})$. Tangents drawn at the ends of chord of length 2 unit of the circle C: $x^2 + y^2 = a^3 - 4b^2 + 12$, meets each other at 'P' on the transverse axis of H.

On the basis of above information, answer the following questions : Number of common tangent of H & C is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (A)

2

Answer 2

The hyperbola is given by $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Let the eccentricity of H be $e$ and the eccentricity of its conjugate hyperbola be $e'$. The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. The eccentricity of H is $e = \sqrt{1 + \frac{b^2}{a^2}}$. The eccentricity of the conjugate hyperbola is $e' = \sqrt{1 + \frac{a^2}{b^2}}$. Given that $e = e'$, we have $\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{a^2}{b^2}}$. Squaring both sides gives $1 + \frac{b^2}{a^2} = 1 + \frac{a^2}{b^2}$, which implies $\frac{b^2}{a^2} = \frac{a^2}{b^2}$. Assuming $a, b \neq 0$, this gives $b^4 = a^4$. Since $a, b$ are positive, $b^2 = a^2$, so $a=b$. The hyperbola is a rectangular hyperbola $x^2 - y^2 = a^2$. The eccentricity of this hyperbola 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 a focus to any tangent of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the auxiliary circle $x^2 + y^2 = a^2$. In our case, the foot of the perpendicular from a focus of $x^2 - y^2 = a^2$ to one of its tangents is given as $(\frac{1}{2}, \frac{\sqrt{15}}{2})$. This point must lie on the auxiliary circle $x^2 + y^2 = a^2$. So, $(\frac{1}{2})^2 + (\frac{\sqrt{15}}{2})^2 = a^2$. $\frac{1}{4} + \frac{15}{4} = a^2$. $\frac{16}{4} = a^2$. $a^2 = 4$. Since $a=b$, $b^2 = 4$. The equation of the hyperbola H is $x^2 - y^2 = 4$.

The equation of the circle C is $x^2 + y^2 = a^3 - 4b^2 + 12$. Substituting $a^2 = 4$ and $b^2 = 4$, and noting $a=2$ (since $a$ is a semi-axis length, it's positive), we get: $R_C^2 = (2)^3 - 4(4) + 12 = 8 - 16 + 12 = 4$. The equation of the circle C is $x^2 + y^2 = 4$. The circle C is centered at the origin with radius $R_C = 2$.

We need to find the number of common tangents to the hyperbola $x^2 - y^2 = 4$ and the circle $x^2 + y^2 = 4$.

Let's check for common tangents of the form $y = mx + c$. The condition for the line $y = mx + c$ to be tangent to the circle $x^2 + y^2 = 4$ is $c^2 = R_C^2(1 + m^2) = 4(1 + m^2) = 4m^2 + 4$. The condition for the line $y = mx + c$ to be tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$. For $x^2 - y^2 = 4$, which is $\frac{x^2}{4} - \frac{y^2}{4} = 1$, we have $a^2 = 4$ and $b^2 = 4$. The condition for tangency to the hyperbola is $c^2 = 4m^2 - 4$.

For a common tangent, the value of $c^2$ must be the same: $4m^2 + 4 = 4m^2 - 4$. $4 = -4$. This is a contradiction, which means there are no common tangents of the form $y = mx + c$.

Now, let's check for vertical tangents of the form $x = k$. For the circle $x^2 + y^2 = 4$, vertical tangents occur at the points where the circle intersects the x-axis, i.e., $(\pm 2, 0)$. The equations of the vertical tangents are $x = \pm 2$. For the hyperbola $x^2 - y^2 = 4$, the vertices are at $(\pm 2, 0)$. The tangents at the vertices are vertical. The tangent at $(x_1, y_1)$ on the hyperbola is $xx_1 - yy_1 = 4$. At the vertex $(2,0)$, the tangent is $x(2) - y(0) = 4$, which is $2x = 4$, or $x = 2$. At the vertex $(-2,0)$, the tangent is $x(-2) - y(0) = 4$, which is $-2x = 4$, or $x = -2$. So, the lines $x=2$ and $x=-2$ are tangents to the hyperbola.

Both $x=2$ and $x=-2$ are tangents to the circle and the hyperbola. These are common tangents. Since we found no common tangents of the form $y=mx+c$ and only two vertical common tangents, the total number of common tangents is 2.

The two curves touch at the points $(\pm 2, 0)$, and the common tangents are the vertical lines $x = \pm 2$ at these points.