This problem is a comprehensive test of several fundamental concepts in coordinate geometry, including tangents to curves, parallel lines, intersection of lines and parabolas, and the distance formula. Our objective is to determine the squared distance between two points, A and B. We will achieve this by systematically finding the coordinates of A and B through a series of well-defined steps.

Here's our plan:

1. Determine the coordinates of Point A: This involves finding the equation of the tangent to the given curve at point P and then finding where this tangent intersects the $y$-axis.
2. Determine the equation of the line passing through P and parallel to $x-3y=6$: This line will be used to find point B.
3. Determine the coordinates of Point B: This requires finding the intersection of the line from step 2 with the parabola $y^2=4x$, and then using the additional condition that B lies on the line $2x-3y=8$ to pinpoint the exact point.
4. Calculate $(\mathrm{AB})^2$: Once A and B are known, we will use the distance formula to find the squared distance between them.

Let's break down each step in detail.

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1. Finding the Equation of the Tangent at P and Coordinates of Point A

Key Concept: Equation of Tangent to a General Quadratic Curve For a general quadratic curve of the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, the equation of the tangent at a point $(x_1, y_1)$ on the curve can be found using the following transformation rules (often called T-substitution):

• Replace $x^2$ with $x x_1$
• Replace $y^2$ with $y y_1$
• Replace $x$ with $\frac{x+x_1}{2}$
• Replace $y$ with $\frac{y+y_1}{2}$
• Replace $xy$ with $\frac{x y_1 + y x_1}{2}$ This method is efficient and avoids differentiation for standard quadratic forms.

Step-by-step Derivation: The given curve is $x^{2}+2 x-4 y+9=0$ The point of tangency is $\mathrm{P}(1,3)$. So, we have $x_1=1$ and $y_1=3$.

Let's apply the transformation rules to each term in the curve's equation:

• For $x^2$: Replace with $x \cdot x_1 = x \cdot 1 = x$.
• For $2x$: Replace with $2 \cdot \frac{x+x_1}{2} = 2 \cdot \frac{x+1}{2} = x+1$.
• For $-4y$: Replace with $-4 \cdot \frac{y+y_1}{2} = -4 \cdot \frac{y+3}{2} = -2(y+3)$.
• The constant term $+9$ remains unchanged.

Now, substitute these transformed terms back into the curve's equation to get the tangent equation: $x + (x+1) - 2(y+3) + 9 = 0$ Simplify the equation: $x + x + 1 - 2y - 6 + 9 = 0$ Combine like terms: $2x - 2y + 4 = 0$ To simplify, divide the entire equation by 2: $x - y + 2 = 0$ This is the equation of the tangent line to the curve $x^2+2x-4y+9=0$ at point $\mathrm{P}(1,3)$.

Finding Point A: Point A is the intersection of this tangent line with the $y$-axis. Any point on the $y$-axis has an $x$-coordinate of 0. To find A, we substitute $x=0$ into the tangent equation $x - y + 2 = 0$: $0 - y + 2 = 0$ $-y = -2$ $y = 2$ Therefore, the coordinates of point A are $\mathrm{A}(0,2)$.

Tip (Alternative Method - Implicit Differentiation): If you prefer, you can also find the tangent using implicit differentiation. Differentiate the curve's equation $x^2+2x-4y+9=0$ with respect to $x$: $\frac{d}{dx}(x^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(4y) + \frac{d}{dx}(9) = 0$ $2x + 2 - 4\frac{dy}{dx} + 0 = 0$ Isolate $\frac{dy}{dx}$: $2x + 2 = 4\frac{dy}{dx}$ $\frac{dy}{dx} = \frac{2x+2}{4} = \frac{x+1}{2}$ The slope of the tangent at point $\mathrm{P}(1,3)$ is obtained by substituting $x=1$ into the derivative: $m = \left.\frac{dy}{dx}\right|_{(1,3)} = \frac{1+1}{2} = \frac{2}{2} = 1$ Now, use the point-slope form of a line, $y-y_1 = m(x-x_1)$, with $\mathrm{P}(1,3)$ and $m=1$: $y - 3 = 1(x - 1)$ $y - 3 = x - 1$ $x - y + 2 = 0$ This confirms our previous result for the tangent line, and consequently, the coordinates of A $(0,2)$.

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2. Finding the Equation of the Line Through P Parallel to $x-3y=6$

Key Concept: Parallel Lines and Slope Two non-vertical lines are parallel if and only if they have the same slope. The slope of a line given in the general form $Ax+By+C=0$ is $m = -\frac{A}{B}$. The slope of a line given in the slope-intercept form $y=mx+c$ is $m$.

Step-by-step Derivation: The line we need to be parallel to is $x-3y=6$. To find its slope, we can rewrite it in slope-intercept form ($y=mx+c$): $3y = x - 6$ $y = \frac{1}{3}x - 2$ The slope of this given line is $m_{\text{given}} = \frac{1}{3}$.

Since the new line passes through $\mathrm{P}(1,3)$ and is parallel to $x-3y=6$, its slope will be the same: $m_{\text{new}} = \frac{1}{3}$. Now, we use the point-slope form of a line, $y-y_1 = m(x-x_1)$, with point $\mathrm{P}(1,3)$ and $m_{\text{new}} = \frac{1}{3}$: $y - 3 = \frac{1}{3}(x - 1)$ To eliminate the fraction and get integer coefficients, multiply both sides by 3: $3(y - 3) = x - 1$ $3y - 9 = x - 1$ Rearrange the terms to the standard form $Ax+By+C=0$: $x - 3y + 8 = 0$ This is the equation of the line passing through P and parallel to $x-3y=6$. Let's call this line $L_1$.

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3. Finding the Coordinates of Point B

Key Concept: Intersection of a Line and a Parabola To find the points where a line intersects a curve (like a parabola), we solve their equations simultaneously. This typically involves substituting an expression for one variable from the linear equation into the curve's equation, resulting in a single-variable quadratic equation. The solutions to this quadratic equation will give the coordinates of the intersection points.

Step-by-step Derivation: The line $L_1$ is $x - 3y + 8 = 0$. The parabola is $y^2 = 4x$. We need to find the point B where $L_1$ intersects the parabola.

From the equation of line $L_1$, it's easiest to express $x$ in terms of $y$: $x = 3y - 8$ Now, substitute this expression for $x$ into the parabola's equation $y^2 = 4x$: $y^2 = 4(3y - 8)$ Expand and rearrange the terms to form a standard quadratic equation in $y$: $y^2 = 12y - 32$ $y^2 - 12y + 32 = 0$ To solve this quadratic equation for $y$, we can factor it. We look for two numbers that multiply to 32 and add up to -12. These numbers are -4 and -8. $(y - 4)(y - 8) = 0$ This gives two possible $y$-coordinates for the intersection points: $y = 4 \quad \text{or} \quad y = 8$

Now, we find the corresponding $x$-coordinates for each $y$ value using $x = 3y - 8$:

• If $y=4$: $x = 3(4) - 8 = 12 - 8 = 4$. This gives a potential intersection point $(4,4)$.
• If $y=8$: $x = 3(8) - 8 = 24 - 8 = 16$. This gives another potential intersection point $(16,8)$.

The problem states that point B lies on the line $2x-3y=8$. This is a crucial additional condition that will allow us to uniquely identify B from the two potential intersection points.

Checking the condition for potential points:

• For point $(4,4)$: Substitute $x=4$ and $y=4$ into the equation $2x-3y=8$: $2(4) - 3(4) = 8 - 12 = -4$ Since $-4 \neq 8$, the point $(4,4)$ does not lie on the line $2x-3y=8$.
• For point $(16,8)$: Substitute $x=16$ and $y=8$ into the equation $2x-3y=8$: $2(16) - 3(8) = 32 - 24 = 8$ Since $8 = 8$, the point $(16,8)$ satisfies the given condition.

Therefore, the coordinates of point B are $\mathrm{B}(16,8)$.

Common Mistake Alert: It is easy to forget about or misinterpret additional conditions given in the problem. Always ensure you use all constraints to select the correct point if multiple candidates arise.

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4. Calculating $(\mathrm{AB})^2$

Key Concept: Distance Formula The squared distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system is given by the formula: $(\text{Distance})^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ We are asked for the squared distance, so we don't need to take the square root.

Step-by-step Calculation: We have the coordinates of point A as $(0,2)$ and point B as $(16,8)$. Let $(x_1, y_1) = (0,2)$ and $(x_2, y_2) = (16,8)$.

Substitute these coordinates into the distance formula: $(\mathrm{AB})^2 = (16 - 0)^2 + (8 - 2)^2$ Perform the subtractions: $(\mathrm{AB})^2 = (16)^2 + (6)^2$ Calculate the squares: $(\mathrm{AB})^2 = 256 + 36$ Add the values: $(\mathrm{AB})^2 = 292$

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Summary and Key Takeaway

This problem was an excellent exercise in integrating various coordinate geometry concepts. We systematically progressed through the following steps:

1. We found the equation of the tangent to a quadratic curve at a given point using the T-substitution method (or implicit differentiation) and identified point A as $(0,2)$.
2. We constructed a line parallel to a given line and passing through point P, yielding the equation $x-3y+8=0$.
3. We determined the intersection points of this line with the parabola $y^2=4x$ and then used the additional condition that B lies on $2x-3y=8$ to uniquely identify point B as $(16,8)$.
4. Finally, we calculated the squared distance between A and B using the distance formula, resulting in $(\mathrm{AB})^2 = 292$.

The problem highlights the importance of:

• Methodical Problem Solving: Breaking down a complex problem into smaller, manageable steps.
• Concept Mastery: Understanding how to apply different geometric formulas and techniques.
• Attention to Detail: Carefully using all given conditions, especially when multiple possibilities arise (like the two intersection points for B).