Question

The equations of a hyperbola and a circle are given below

$H:(x+2)^2 - (y-3)^2 = -2$ $C:(x-2)^2+(y-3)^2 = 10$

Let circle 'C' intersects the line y = 3, at points P and Q, such that P is closer to (-2, 3).

A line L passing through P meets the hyperbola H and the circle C at a total of 3 points.

Then, number of such straight lines L, is equal to ____

Answer 1

2

Answer 2

Solution:

1. The circle is

   $$(x-2)^2 + (y-3)^2 = 10.$$

   Its intersection with the horizontal line $y=3$ gives

   $$(x-2)^2 = 10 \quad\Longrightarrow\quad x = 2\pm\sqrt{10}.$$

   Since the point closer to $(-2,3)$ will have the smaller $x$-value, we get

   $$P=(2-\sqrt{10},3).$$

2. Let a line $L$ through $P$ be

   $$y-3 = m\Big(x-(2-\sqrt{10})\Big).$$

   This line meets the circle $C$ at $P$ and (generically) at another point $Q$.

3. For the union of intersections with the hyperbola and circle to total 3 distinct points, the line must intersect one of the conics in a double (tangent) manner. Since $P$ is fixed on the circle and is not on the hyperbola, the only possibility is to have the line tangent to the hyperbola

   $$(x+2)^2 - (y-3)^2 = -2,$$

   so that $L$ and $H$ meet in a single (double) point while $L$ meets $C$ in two distinct points (namely, $P$ and $Q$).

4. Substitute $y-3 = m\bigl(x-(2-\sqrt{10})\bigr)$ into $H$:

   $$(x+2)^2 - \Bigl[m\Bigl(x-(2-\sqrt{10})\Bigr)\Bigr]^2 = -2.$$

   Rewriting,

   $$(x+2)^2 - m^2\Bigl(x-(2-\sqrt{10})\Bigr)^2 + 2=0.$$

   This is a quadratic in $x$. For the line to be tangent to $H$ the discriminant must be zero.

5. Writing the quadratic in the form

   $$A x^2 + B x + C = 0,$$

   after expanding we obtain:

   $$(1-m^2)x^2 + \Bigl[4+ 2m^2(2-\sqrt{10})\Bigr]x + \Bigl[6 - m^2(2-\sqrt{10})^2\Bigr]=0.$$

   Tangency requires

   $$\Bigl[4 + 2m^2(2-\sqrt{10})\Bigr]^2 - 4(1-m^2)\Bigl[6-m^2(2-\sqrt{10})^2\Bigr] = 0.$$

6. Simplify by letting $\alpha = 2-\sqrt{10}$. Then,

   $$[4+2m^2\alpha]^2 - 4(1-m^2)[6-m^2\alpha^2] = 0.$$

   Factor $2$ from the first bracket:

   $$4(2+m^2\alpha)^2 - 4(1-m^2)[6-m^2\alpha^2] = 0.$$

   Dividing by 4 and expanding yields:

   $$(2+m^2\alpha)^2 - (1-m^2)(6-m^2\alpha^2) = 0.$$

   Expanding further,

   $$4 + 4m^2\alpha + m^4\alpha^2 - \Bigl[6-6m^2-m^2\alpha^2+m^4\alpha^2\Bigr]=0.$$

   The $m^4\alpha^2$ terms cancel and we get:

   $$4 + 4m^2\alpha + 6m^2 + m^2\alpha^2 - 6 = 0.$$

   That simplifies to:

   $$m^2\Bigl(\alpha^2+4\alpha+6\Bigr) = 2.$$

7. Now compute $\alpha^2$:

   $$\alpha^2 = (2-\sqrt{10})^2 = 4-4\sqrt{10}+10 = 14-4\sqrt{10}.$$

   Thus,

   $$\alpha^2+4\alpha+6 = (14-4\sqrt{10}) + 4(2-\sqrt{10}) + 6 = 14-4\sqrt{10}+8-4\sqrt{10}+6 = 28 - 8\sqrt{10}.$$

   So,

   $$m^2 = \frac{2}{28-8\sqrt{10}} = \frac{2}{4(7-2\sqrt{10})} = \frac{1}{2(7-2\sqrt{10})}.$$

8. Since $m^2$ is positive, there are 2 possible real values of $m$ (namely $m=\pm \frac{1}{\sqrt{2(7-2\sqrt{10})}}$). Hence, there are 2 lines $L$ which satisfy the given condition.

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Explanation (minimal):

• Find $P=(2-\sqrt{10},3)$ from circle $C$ along $y=3$.
• A line through $P$: $y-3=m(x-(2-\sqrt{10}))$ meets $C$ in $P$ and $Q$.
• To have a total of 3 intersections with $C$ and hyperbola $H$, $L$ must be tangent to $H$.
• Substitute the line in $H$, set the quadratic discriminant to zero, simplify to get: $$m^2(\alpha^2+4\alpha+6)=2,\quad \alpha=2-\sqrt{10}.$$
• Compute $\alpha^2+4\alpha+6=28-8\sqrt{10}$ and solve to get two possible slopes.
• Thus, the number of such lines $L$ is 2.