Let a conic C pass through the point (4, -2) and P(x, y), x ≥ 3, be any point on C. Let the slope of

Q: Wait, why is the slope of the tangent just $\frac{dy}{dx}$? Isn't that too simple?

That is a brilliant question! In the context of a curve defined by $y = f(x)$, the slope of the tangent line at any point $P(x, y)$ is indeed defined as the derivative $\frac{dy}{dx}$. It is the fundamental bridge between the geometry of the curve and the calculus of its rate of change. You are not oversimplifying; you are using the definition correctly!

Q: How do I know this is a parabola and not an ellipse or hyperbola?

Great observation. Once you integrate and simplify the equation to $(y + 5)^2 = 9(x - 3)$, you can compare it to the standard form of a parabola, $(y - k)^2 = 4a(x - h)$. Because only one variable is squared, it must be a parabola. If both $x$ and $y$ were squared with the same sign, it could be a circle or ellipse; if they had opposite signs, it would be a hyperbola.

Q: Is the focal distance always $x_1 - h + a$?

Yes, for a parabola opening to the right, the distance from any point $(x_1, y_1)$ on the parabola to the focus is given by $x_1 + a - h$. This is derived from the definition of a parabola: the distance from any point on the curve to the focus is equal to its distance to the directrix. It is a very powerful property to keep in your toolkit!

Question

Let a conic $C$ pass through the point $(4, -2)$ and $P(x, y), x \ge 3$, 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 ______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a curve, a conic section $C$. You are at a point $P(x, y)$, and there is a fixed point $V$ located at $(3, -5)$ that acts as an anchor for our problem.

The problem states that the slope of the tangent at $P$ is exactly half the slope of the line segment $VP$.

The slope of the tangent at $P$ is the derivative $\frac{dy}{dx}$. The slope of the line joining $P(x, y)$ and $V(3, -5)$ is given by the formula $\frac{y - (-5)}{x - 3} = \frac{y + 5}{x - 3}$.

The governing condition is:

$$
\frac{dy}{dx} = \frac{1}{2} \left( \frac{y + 5}{x - 3} ight)
$$

This differential equation is the heartbeat of the problem, capturing the geometric essence of the curve.

### The Art of Integration

To solve this, we use the method of separation of variables. We group all $y$ terms with $dy$ and all $x$ terms with $dx$:

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

Now, we integrate both sides:

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

Performing the integration, we obtain:

$$
2 \ln|y + 5| = \ln|x - 3| + C
$$

### Determining the Constant

We need the specific curve that passes through $(4, -2)$. Substituting these coordinates allows us to solve for the constant $C$:

$$
2 \ln|-2 + 5| = \ln|4 - 3| + C \Rightarrow 2 \ln(3) = \ln(1) + C
$$

Since $\ln(1) = 0$, we find $C = 2 \ln(3) = \ln(9)$.

### Unveiling the Parabola

Substituting $C$ back into our equation, we get:

$$
2 \ln(y + 5) = \ln(x - 3) + \ln(9)
$$

Using the properties of logarithms, this simplifies to:

$$
\ln((y + 5)^2) = \ln(9(x - 3)) \Rightarrow (y + 5)^2 = 9(x - 3)
$$

This is the equation of a parabola. By comparing it to the standard form $(y - k)^2 = 4a(x - h)$, we identify the vertex $(h, k) = (3, -5)$ and the focal length $4a = 9$, so $a = \frac{9}{4}$.

### Final Calculation

The focal distance $d$ of a point $(x_1, y_1)$ on this parabola is the distance from the point to the focus. For this parabola, the formula is $d = (x_1 - h) + a$.

For the point $(7, 1)$, we have $x_1 = 7$, $h = 3$, and $a = \frac{9}{4}$:

$$
d = (7 - 3) + \frac{9}{4} = 4 + \frac{9}{4} = \frac{25}{4}
$$

The question asks for the value of $12d$:

$$
12d = 12 	imes \frac{25}{4} = 3 	imes 25 = 75
$$

Through the power of calculus and geometric insight, we have arrived at our final answer: **75**.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem Setup

Defining the Slopes

Setting Up the Differential Equation

Separating the Variables

Integrating Both Sides

Finding the Constant $C$

Evaluating the Constant $C$

Simplifying the Equation

Identifying the Conic

Understanding Focal Distance

Calculating $d$ for $(7, 1)$

Evaluating the Focal Distance $d$

Final Answer: $12 d$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Integration

Determining the Constant

Unveiling the Parabola

Final Calculation

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