6.5 Q-3

Question Statement

Prove that the locus of points $P(x, y)$, such that the sum of distances from $P$ to two fixed points $F(-c, 0)$ and $F'(c, 0)$ is constant, i.e., $|\text{PF}| + |\text{PF}'| = 2a$, forms an ellipse.

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Background and Explanation

An ellipse is a set of points where the sum of distances from any point on the ellipse to two fixed points (foci) is constant. This geometric property defines the ellipse mathematically. For an ellipse:

• $a$ is the semi-major axis.
• $c$ is the distance from the center to each focus.
• $b$ is the semi-minor axis, and it satisfies $b^2 = a^2 - c^2$.

The equation of the ellipse in standard form is:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

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Solution

Let the two foci of the ellipse be $F(-c, 0)$ and $F'(c, 0)$. The given condition is:

$$|\text{PF}| + |\text{PF}'| = 2a$$

Step 1: Express the distances
For a point $P(x, y)$:

$$|\text{PF}| = \sqrt{(x + c)^2 + y^2}, \quad |\text{PF}'| = \sqrt{(x - c)^2 + y^2}$$

Substituting into the condition:

$$\sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = 2a$$

Step 2: Square both sides
Square both sides to eliminate the square roots:

$$\sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = 2a$$ $$(x + c)^2 + y^2 + (x - c)^2 + y^2 + 2 \sqrt{{(x + c)^2 + y^2}{(x - c)^2 + y^2}} = 4a^2$$

Step 3: Expand and simplify
Simplify the left-hand side:

$$x^2 + c^2 + 2xc + y^2 + x^2 + c^2 - 2xc + y^2 + 2 \sqrt{((x + c)^2 + y^2)((x - c)^2 + y^2)} = 4a^2$$ $$2x^2 + 2y^2 + 2c^2 + 2 \sqrt{((x + c)^2 + y^2)((x - c)^2 + y^2)} = 4a^2$$

Step 4: Isolate the square root
Rearrange to isolate the square root term:

$$\sqrt{((x + c)^2 + y^2)((x - c)^2 + y^2)} = 2a^2 - (x^2 + y^2 + c^2)$$

Step 5: Square again
Square both sides to eliminate the square root:

$$((x + c)^2 + y^2)((x - c)^2 + y^2) = (2a^2 - (x^2 + y^2 + c^2))^2$$

Expanding and simplifying leads to:

$$(a^2 - c^2)x^2 + a^2y^2 = a^2(a^2 - c^2)$$

Step 6: Rewrite in standard form
Divide through by $a^2(a^2 - c^2)$:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad \text{where } b^2 = a^2 - c^2$$

Thus, the locus of $P(x, y)$ satisfies the equation of an ellipse.

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Key Formulas or Methods Used

1. Ellipse Definition: The sum of distances to the two foci is constant.
2. Distance Formula:

$$|\text{PF}| = \sqrt{(x + c)^2 + y^2}, \quad |\text{PF}'| = \sqrt{(x - c)^2 + y^2}$$

3. Ellipse Standard Equation:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad \text{where } b^2 = a^2 - c^2$$

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Summary of Steps

1. Express the distances $|\text{PF}|$ and $|\text{PF}'|$.
2. Use the given condition $|\text{PF}| + |\text{PF}'| = 2a$.
3. Square both sides to eliminate square roots, simplify, and isolate terms.
4. Derive the standard form of the ellipse equation.

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