Parabola from Focus and Directrix

 

❓ Question

Let $P$ be the parabola whose focus is

$$(-2,\,1)$$

and whose directrix is

$$2x + y + 2 = 0.$$

Find the sum of the ordinates of the points on $P$ whose abscissa is

$$x=-2.$$

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🖼️ Question Image

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✍️ Short Solution

A parabola is defined as the locus of a point whose distance from the focus equals its perpendicular distance from the directrix.

We will:
1️⃣ Write the focus–directrix condition
2️⃣ Substitute $x = -2$
3️⃣ Solve for $y$
4️⃣ Add the ordinates

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🔹 Step 1 — Use focus–directrix definition

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

Distance from focus $(-2,1)$:

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

Distance from directrix $2x + y + 2 = 0$:

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

Equating squares:

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

Multiply both sides by 5:

$$5[(x+2)^2 + (y-1)^2] = (2x + y + 2)^2$$

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🔹 Step 2 — Substitute $x = -2$

Since we are asked about points whose abscissa is $-2$, put $x=-2$:

Left side:

$$5[(0)^2 + (y-1)^2] = 5(y-1)^2$$

Right side:

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

So the equation becomes:

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

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🔹 Step 3 — Solve for $y$

Expand both sides:

$$5(y^2 - 2y + 1) = y^2 - 4y + 4$$
$$5y^2 - 10y + 5 = y^2 - 4y + 4$$

Bring all terms to one side:

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

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🔹 Step 4 — Find the sum of ordinates

For the quadratic:

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

Sum of roots:

$$\text{Sum of } y = \frac{6}{4} = \frac{3}{2}$$

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✅ Final Answer

$$\boxed{\frac{3}{2}}$$

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