Let P be the parabola, whose focus is (-2, 1) and directrix is 2x + y + 2 = 0. Then the sum of the o

Q: Wait, why do we use $PS^2 = PM^2$ instead of just $PS = PM$?

Great question! While $PS = PM$ is the definition, it involves square roots on both sides. Squaring both sides eliminates the radicals, making the algebra much cleaner and less prone to sign errors. It is a standard tactical move in coordinate geometry.

Q: Do I really need to find the individual values of $y$ to get the sum?

Not at all! This is the beauty of Vieta's formulas. Once you have the quadratic equation in $y$, the sum of the roots is directly available from the coefficients. Finding the roots individually is extra work you don't need to do.

Q: Why is the distance to the directrix divided by $\sqrt{2^2 + 1^2}$?

That is the standard formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$, which is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Here, $A=2$ and $B=1$.

Q: What if the line $x = -2$ was the directrix itself?

If the line $x = -2$ were the directrix, the parabola would be degenerate or the geometry would change entirely. However, here $x = -2$ is just a vertical line intersecting the parabola, which is perfectly valid.

Q: Is it possible for the sum of ordinates to be negative?

Absolutely! The sum of roots depends on the coefficients of your quadratic equation. In this specific problem, the geometry leads to a positive sum, but always trust your algebraic derivation over intuition.

Question

Let $P$ be the parabola, whose focus is $(-2, 1)$ and directrix is $2x + y + 2 = 0$. Then the sum of the ordinates of the points on $P$, whose abscissa is $-2$, is

Options:

(A) $\frac{3}{2}$

(B) $\frac{5}{2}$

(C) $\frac{1}{4}$

(D) $\frac{3}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty coordinate plane. You have a single point, a beacon of light at $S(-2, 1)$, which we call the focus. You also have a rigid, infinite boundary, a line defined by $L: 2x + y + 2 = 0$, which we call the directrix.

A parabola is the locus of a point $P(x, y)$ that maintains a perfect, unwavering balance: its distance from the focus $S$ must always equal its perpendicular distance from the directrix $L$. This fundamental definition, $PS = PM$, is the key to unlocking the secrets of conic sections.

### The Algebraic Bridge

To translate this geometric dance into the language of algebra, we use the distance formula. The distance from $P(x, y)$ to the focus $S(-2, 1)$ is:

$$PS = \sqrt{(x + 2)^2 + (y - 1)^2}$$

The perpendicular distance from $P(x, y)$ to the line $2x + y + 2 = 0$ is given by:

$$PM = \frac{|2x + y + 2|}{\sqrt{2^2 + 1^2}} = \frac{|2x + y + 2|}{\sqrt{5}}$$

Equating $PS = PM$ and squaring both sides to eliminate the radicals, we obtain the general equation of the parabola:

$$(x + 2)^2 + (y - 1)^2 = \frac{(2x + y + 2)^2}{5}$$

### The Intersection of Paths

To find the points on this parabola where the abscissa is $x = -2$, we substitute $x = -2$ directly into our squared equation. The left side simplifies as follows:

$$(-2 + 2)^2 + (y - 1)^2 = (y - 1)^2$$

The right side simplifies as follows:

$$\frac{(2(-2) + y + 2)^2}{5} = \frac{(y - 2)^2}{5}$$

Equating these results, we arrive at the following relationship:

$$(y - 1)^2 = \frac{(y - 2)^2}{5}$$

### The Final Symmetry

Multiplying by 5, we get $5(y - 1)^2 = (y - 2)^2$. Expanding both sides yields:

$$5(y^2 - 2y + 1) = y^2 - 4y + 4$$

$$5y^2 - 10y + 5 = y^2 - 4y + 4$$

Rearranging all terms to one side, we arrive at the quadratic equation:

$$4y^2 - 6y + 1 = 0$$

We are looking for the sum of the ordinates, which are the roots of this quadratic equation. By Vieta's formulas, for a quadratic $ay^2 + by + c = 0$, the sum of the roots is given by $-\frac{b}{a}$.

Here, $a = 4$ and $b = -6$. Therefore, the sum of the ordinates is:

$$	ext{Sum} = -\left(\frac{-6}{4}ight) = \frac{6}{4} = \mathbf{\frac{3}{2}}$$

Animated Solution for Mathematics - Conic Sections: Let $P$ be the parabola, whose focus is $(- 2, 1)$ and directrix is $2 x + y + 2 = 0$ . Then the sum of the ordinates of the points on $P$ , whose abscissa is $- 2$ , is

Select Answer:

Visualized Solution

Visualize the Given Data

The Locus Definition

Formulating the Distances

Squaring to get the Equation

Applying the Condition $x = - 2$

Substituting $x = - 2$

Simplifying the Left Hand Side

Simplifying the Right Hand Side

Equating and Expanding

Forming the Quadratic Equation

Finding the Sum of Ordinates

Final Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Algebraic Bridge

The Intersection of Paths

The Final Symmetry

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