Introduction: Understanding the Parabola's Definition

A parabola is a fundamental conic section defined by a unique geometric property: it is the locus of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed straight line (called the directrix).

If we denote an arbitrary point on the parabola as $P(x,y)$, the focus as $S(x_F, y_F)$, and the directrix as the line $L: Ax+By+C=0$, then the defining property of the parabola is: $PS = PM$ where $PS$ is the distance from point $P$ to the focus $S$, and $PM$ is the perpendicular distance from point $P$ to the directrix $L$. This core concept will be our starting point for deriving the parabola's equation.

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Step 1: Formulating the Equation of the Parabola

Our primary goal here is to translate the given geometric information (focus and directrix) into an algebraic equation representing the parabola.

Given Information:

• Focus $S = (-2, 1)$
• Directrix $L: 2x + y + 2 = 0$

Let $P(x,y)$ be any point on the parabola.

1. Calculate the Distance from $P$ to the Focus ($PS$): We use the standard distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. For $P(x,y)$ and $S(-2,1)$: $PS = \sqrt{(x - (-2))^2 + (y - 1)^2} = \sqrt{(x+2)^2 + (y-1)^2}$ Why this step? This expression represents the distance of any point on the parabola from the focus, which is one part of our defining equality $PS=PM$.

2. Calculate the Perpendicular Distance from $P$ to the Directrix ($PM$): We use the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$, which is $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. For point $P(x,y)$ and directrix $2x+y+2=0$ (where $A=2, B=1, C=2$): $PM = \frac{|2x+y+2|}{\sqrt{2^2+1^2}} = \frac{|2x+y+2|}{\sqrt{4+1}} = \frac{|2x+y+2|}{\sqrt{5}}$ Why this step? This expression represents the perpendicular distance of any point on the parabola from the directrix, completing the other part of our defining equality $PS=PM$.

3. Equate $PS$ and $PM$ to form the Parabola's Equation: According to the definition of a parabola, $PS = PM$. Equating the expressions we found: $\sqrt{(x+2)^2 + (y-1)^2} = \frac{|2x+y+2|}{\sqrt{5}}$ Why this step? This is the algebraic representation of the parabola based on its fundamental geometric definition. This equation holds true for every point $(x,y)$ that lies on the parabola. Tip: Typically, to get the standard form of the parabola's equation, one would square both sides at this point. However, for this specific problem, we are looking for points with a fixed $x$-coordinate, so substituting $x$ first might simplify the algebra.

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Step 2: Substituting the Given Abscissa (x-coordinate)

The problem asks for the sum of the ordinates (y-coordinates) of points on the parabola whose abscissa (x-coordinate) is $-2$. To find these specific points, we substitute $x = -2$ into the parabola's equation derived in Step 1.

Why this step? We are not interested in the general equation of the parabola, but rather in specific points on it where $x=-2$. Substituting this value will reduce the equation to an expression solely in terms of $y$, allowing us to solve for the ordinates.

Substituting $x = -2$ into the equation $\sqrt{(x+2)^2 + (y-1)^2} = \frac{|2x+y+2|}{\sqrt{5}}$: $\sqrt{(-2+2)^2 + (y-1)^2} = \frac{|2(-2)+y+2|}{\sqrt{5}}$ Simplify the terms: $\sqrt{0^2 + (y-1)^2} = \frac{|-4+y+2|}{\sqrt{5}}$ $\sqrt{(y-1)^2} = \frac{|y-2|}{\sqrt{5}}$

Crucial Note on Square Roots and Absolute Values: When simplifying $\sqrt{A^2}$, it is absolutely vital to remember that $\sqrt{A^2} = |A|$, and not just $A$. This is because the square root symbol $\sqrt{}$ denotes the principal (non-negative) square root. For example, $\sqrt{(-3)^2} = \sqrt{9} = 3$, which is $|-3|$. If we simplified it to $-3$, it would be incorrect. Therefore, $\sqrt{(y-1)^2}$ simplifies to $|y-1|$.

The equation now becomes: $|y-1| = \frac{|y-2|}{\sqrt{5}}$ This equation now contains only $y$, and its solutions will be the ordinates of the points on the parabola at $x=-2$.

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Step 3: Solving for the Ordinates (y-values)

Now we need to solve the absolute value equation obtained in Step 2 for $y$.

Why this step? The equation now contains only $y$. Solving it will give us the specific $y$-coordinates (ordinates) of the points on the parabola where $x=-2$.

To eliminate the absolute values and the square root from the denominator, we square both sides of the equation: $\left(|y-1|\right)^2 = \left(\frac{|y-2|}{\sqrt{5}}\right)^2$ Recall that for any real number $A$, $|A|^2 = A^2$. So, this simplifies to: $(y-1)^2 = \frac{(y-2)^2}{5}$

Now, we expand both sides and rearrange the terms to form a standard quadratic equation: $5(y-1)^2 = (y-2)^2$ $5(y^2 - 2y + 1) = y^2 - 4y + 4$ $5y^2 - 10y + 5 = y^2 - 4y + 4$

Move all terms to one side to form a quadratic equation of the form $Ay^2+By+C=0$: $5y^2 - y^2 - 10y + 4y + 5 - 4 = 0$ $4y^2 - 6y + 1 = 0$ Why this step? We've transformed the equation into a standard quadratic form. The roots of this quadratic equation are the ordinates ($y_1, y_2$) of the points on the parabola where $x=-2$.

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Step 4: Calculating the Sum of the Ordinates

The question asks for the sum of these ordinates ($y_1 + y_2$).

Why this step? Instead of finding the individual roots using the quadratic formula, which would involve radicals, we can directly find their sum using Vieta's formulas. This is a much more efficient and elegant approach for this type of question.

Vieta's Formulas for Quadratic Equations: For a quadratic equation of the form $Ay^2+By+C=0$, if $y_1$ and $y_2$ are its roots, then:

• Sum of roots: $y_1 + y_2 = -\frac{B}{A}$
• Product of roots: $y_1 \cdot y_2 = \frac{C}{A}$

In our quadratic equation, $4y^2 - 6y + 1 = 0$:

• Coefficient $A = 4$
• Coefficient $B = -6$
• Coefficient $C = 1$

Therefore, the sum of the ordinates ($y_1 + y_2$) is: $\text{Sum of ordinates} = -\frac{B}{A} = -\frac{(-6)}{4} = \frac{6}{4} = \frac{3}{2}$

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Summary and Key Takeaway

We have successfully determined the sum of the ordinates of the points on the parabola where the abscissa is $-2$. The solution process involved a systematic application of core concepts:

1. Definition of a Parabola: Using $PS=PM$ to set up the initial equation.
2. Algebraic Manipulation: Substituting the given $x$-coordinate, carefully handling absolute values, and squaring both sides to simplify the equation.
3. Quadratic Equations and Vieta's Formulas: Recognizing that the resulting equation was quadratic in $y$ and efficiently finding the sum of its roots using Vieta's formulas.

This problem highlights the importance of mastering the fundamental definition of conic sections and being adept at algebraic manipulation, especially with square roots and absolute values.

The final answer is $\frac{3}{2}$.

The correct option is (B).