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.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: