Question

Let P, Q, R, S be the points of intersection of the circle $x^2 + y^2 = 4$ and the hyperbola $xy = \sqrt{3}$. If P = $(\alpha,\beta)$ and $\alpha>\beta>0$, then the equation of the tangent drawn at P to the hyperbola is

• A. $x+y=2$
• B. $x+\sqrt{3}y=2\sqrt{3}$
• C. $\sqrt{3}x+y=\sqrt{3}$
• D. $x-y=0$

Hint:

The equation of the tangent to a rectangular hyperbola $xy=c^2$ at a point $(x_1, y_1)$ can be remembered as $\frac{x}{x_1} + \frac{y}{y_1} = 2$, or more commonly as $xy_1+yx_1=2c^2$.

Answer 1

The Correct Option is B

We need to find the coordinates of the intersection point P$(\alpha,\beta)$ in the first quadrant.
Circle: $x^2 + y^2 = 4$.
Hyperbola: $xy = \sqrt{3} \implies y = \sqrt{3}/x$.
Substitute $y$ from the hyperbola equation into the circle equation:
$x^2 + (\frac{\sqrt{3}}{x})^2 = 4 \implies x^2 + \frac{3}{x^2} = 4$.
Multiply by $x^2$: $x^4 + 3 = 4x^2 \implies x^4 - 4x^2 + 3 = 0$.
This is a quadratic equation in $x^2$. Let $z=x^2$.
$z^2 - 4z + 3 = 0 \implies (z-1)(z-3)=0$.
So, $z=1$ or $z=3$. This means $x^2=1$ or $x^2=3$.
If $x^2=1$, then $y^2 = 4-x^2 = 3$. This gives $x=\pm 1, y=\pm\sqrt{3}$. If $x^2=3$, then $y^2 = 4-x^2 = 1$. This gives $x=\pm \sqrt{3}, y=\pm 1$.
The intersection points are $(\pm 1, \pm\sqrt{3})$ and $(\pm \sqrt{3}, \pm 1)$ such that $xy=\sqrt{3}$. The points are $(1,\sqrt{3}), (-1,-\sqrt{3}), (\sqrt{3},1), (-\sqrt{3},-1)$.
We are given that P is $(\alpha, \beta)$ with $\alpha>\beta>0$. Comparing the two possible points in the first quadrant, $(1, \sqrt{3})$ and $(\sqrt{3}, 1)$: For $(1, \sqrt{3})$, we have $\alpha=1, \beta=\sqrt{3}$. Here $\alpha<\beta$. For $(\sqrt{3}, 1)$, we have $\alpha=\sqrt{3}, \beta=1$. Here $\alpha = \sqrt{3} \approx 1.732>\beta=1>0$. This matches the condition. So, the point P is $(\sqrt{3}, 1)$.
The equation of the tangent to the hyperbola $xy=c$ at the point $(x_1, y_1)$ is $xy_1+yx_1=2c$.
Here, $c=\sqrt{3}$ and $(x_1, y_1) = (\sqrt{3}, 1)$.
The tangent equation is $x(1) + y(\sqrt{3}) = 2\sqrt{3}$.
$x + \sqrt{3}y = 2\sqrt{3}$.