Question

Let a conic $C$ pass through the point $(4, -2)$ and $P(x, y), \, x \geq 3$, be any point on $C$. Let the slope of the line touching the conic $C$ only at a single point $P$ be half the slope of the line joining the points $P$ and $(3, -5)$. If the focal distance of the point $(7, 1)$ on $C$ is $d$, then $12d$ equals ______.

Answer 1

Given $P(x, y)$ and $x \geq 3$, the slope of the tangent at $P(x, y)$ to the conic is:

$\frac{dy}{dx} = \frac{1}{2} \frac{y + 5}{x - 3}.$

Rewriting:

$2 \frac{dy}{y + 5} = \frac{1}{x - 3} dx.$

Integrating both sides:

$2 \ln(y + 5) = \ln(x - 3) + C.$

Simplifying:

$\ln(y + 5)^2 = \ln(x - 3) + C,$ $(y + 5)^2 = k(x - 3), \quad \text{where } k = e^C.$

Since the conic passes through $(4, -2)$, substitute:

$(-2 + 5)^2 = k(4 - 3),$ $9 = k(1) \implies k = 9.$

Thus, the conic equation becomes:

$(y + 5)^2 = 9(x - 3).$

This represents a parabola with:

$4a = 9 \implies a = \frac{9}{4}.$

The focal distance $d$ of the point $(7, 1)$ is given by:

$d = \sqrt{\left(\frac{7}{4}\right)^2 + 6^2}.$

Simplifying:

$d = \sqrt{\frac{49}{16} + 36} = \sqrt{\frac{625}{16}} = \frac{25}{4}.$

Thus:

$12d = 12 \times \frac{25}{4} = 75.$

Final Answer: 75.