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

Correct Answer: 75

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.