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

We are given a relationship between the slope of the tangent to a conic \( C \) at a point \( P(x, y) \) and the slope of the line segment joining \( P \) to the point \( (3, -5) \). The conic passes through \( (4, -2) \). We need to find the equation of the conic, determine its properties, and then calculate \( 12d \), where \( d \) is the focal distance of the point \( (7, 1) \) on the conic.

Concept Used:

1. Differential Equations: The slope of the tangent to a curve at a point \( (x, y) \) is given by the derivative \( \frac{dy}{dx} \). The problem statement provides a relation for this derivative, which forms a separable differential equation.

2. Equation of a Parabola: The standard equation of a parabola with vertex at \( (\alpha, \beta) \) and a horizontal axis of symmetry is \( (y - \beta)^2 = 4a(x - \alpha) \). The focus is at \( (\alpha + a, \beta) \) and the directrix is the line \( x = \alpha - a \).

3. Focal Distance of a Parabola: The focal distance of any point \( P(x_0, y_0) \) on a parabola is its distance from the focus. By the definition of a parabola, this distance is also equal to the perpendicular distance from the point \( P \) to the directrix.

Step-by-Step Solution:

Step 1: Formulate the differential equation from the given information.

The slope of the tangent at \( P(x, y) \) is \( m_{tan} = \frac{dy}{dx} \).

The slope of the line joining \( P(x, y) \) and \( Q(3, -5) \) is \( m_{PQ} = \frac{y - (-5)}{x - 3} = \frac{y+5}{x-3} \).

According to the problem, \( m_{tan} = \frac{1}{2} m_{PQ} \). Therefore:

\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{y+5}{x-3} \right) \]

Step 2: Solve this differential equation by separating the variables.

\[ \frac{dy}{y+5} = \frac{1}{2} \frac{dx}{x-3} \]

Step 3: Integrate both sides of the separated equation.

\[ \int \frac{1}{y+5} \, dy = \frac{1}{2} \int \frac{1}{x-3} \, dx \] \[ \ln|y+5| = \frac{1}{2} \ln|x-3| + C \]

Using the property of logarithms \( n \ln a = \ln a^n \):

\[ \ln|y+5| = \ln\sqrt{|x-3|} + C \]

Exponentiating both sides:

\[ |y+5| = e^C \sqrt{|x-3|} \]

Since \( x \geq 3 \), \( |x-3| = x-3 \). Let \( A = \pm e^C \) be a new constant. Squaring both sides gives:

\[ (y+5)^2 = A^2 (x-3) \]

Let \( k = A^2 \). The equation of the family of conics is \( (y+5)^2 = k(x-3) \).

Step 4: Find the value of the constant \( k \) using the fact that the conic passes through the point \( (4, -2) \).

Substitute \( x = 4 \) and \( y = -2 \) into the equation:

\[ (-2 + 5)^2 = k(4 - 3) \] \[ (3)^2 = k(1) \implies k = 9 \]

So, the equation of the conic \( C \) is \( (y+5)^2 = 9(x-3) \).

Step 5: Identify the conic and its parameters.

The equation \( (y+5)^2 = 9(x-3) \) is in the standard form of a parabola, \( (y-\beta)^2 = 4a(x-\alpha) \), with a horizontal axis.

By comparing the equations, we have:

• Vertex \( (\alpha, \beta) = (3, -5) \)
• \( 4a = 9 \implies a = \frac{9}{4} \)

Step 6: Find the equation of the directrix of the parabola.

The equation of the directrix for this type of parabola is \( x = \alpha - a \).

\[ x = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4} \]

The equation of the directrix is \( x - \frac{3}{4} = 0 \) or \( 4x - 3 = 0 \).

Step 7: Calculate the focal distance \( d \) of the point \( (7, 1) \).

The focal distance \( d \) of a point on a parabola is the perpendicular distance from that point to the directrix.

The distance from the point \( (x_0, y_0) = (7, 1) \) to the line \( x - \frac{3}{4} = 0 \) is:

\[ d = \frac{|x_0 - \frac{3}{4}|}{\sqrt{1^2 + 0^2}} = \left| 7 - \frac{3}{4} \right| = \left| \frac{28 - 3}{4} \right| = \frac{25}{4} \]

Final Computation & Result:

Step 8: Calculate the final required value, \( 12d \).

\[ 12d = 12 \times \frac{25}{4} \] \[ 12d = 3 \times 25 = 75 \]

Thus, the value of \( 12d \) is 75.

Answer 2

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.