6.6 Q-3

Question Statement

Let $0 < a < c$, and let $F(-c, 0)$ and $F'(c, 0)$ be two fixed points. Show that the set of points $P(x, y)$ such that $|PF| - |PF'| = \pm 2a$ forms a hyperbola with the equation:

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

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Background and Explanation

A hyperbola is defined as the locus of points where the absolute difference of distances to two fixed points (foci) is constant. This property is captured mathematically in the given problem. By squaring and rearranging terms, we aim to demonstrate that this condition translates into the standard equation of a hyperbola.

Key components:

1. Foci: The two fixed points $(-c, 0)$ and $(c, 0)$.
2. Vertices: Points where the hyperbola intersects its transverse axis.
3. Relation between parameters: $c^2 = a^2 + b^2$, where $b$ is derived from the transverse and conjugate axes.

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Solution

Step 1: Express the condition mathematically

The condition $|PF| - |PF'| = \pm 2a$ translates to:

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

Step 2: Simplify the equation

Squaring both sides to eliminate the square root:

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

Expanding both sides:

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

Combine like terms:

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

Step 3: Isolate the square root

Rearranging:

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

Step 4: Square again

Squaring both sides to eliminate the remaining square root:

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

Expand both sides:

1. For the left-hand side:

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

Expanding gives:

$$(x^2 + y^2 + c^2)^2 - (2xc)^2$$

2. For the right-hand side:

$$\left(x^2 + y^2 + c^2 - 2a^2\right)^2 = (x^2 + y^2 + c^2)^2 - 4a^2(x^2 + y^2 + c^2) + 4a^4$$

Equating both sides:

$$(x^2 + y^2 + c^2)^2 - 4x^2c^2 = (x^2 + y^2 + c^2)^2 - 4a^2(x^2 + y^2 + c^2) + 4a^4$$

Step 5: Simplify and isolate terms

Cancel $(x^2 + y^2 + c^2)^2$ from both sides:

$$-4x^2c^2 = -4a^2(x^2 + y^2 + c^2) + 4a^4$$

Divide by $-4$:

$$x^2c^2 = a^2(x^2 + y^2 + c^2) - a^4$$

Expand and rearrange:

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

Step 6: Normalize the equation

Divide through by $a^2(c^2 - a^2)$:

$$\frac{x^2}{a^2} - \frac{y^2}{c^2 - a^2} = 1$$

This is the standard equation of a hyperbola, where:

$$b^2 = c^2 - a^2$$

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Key Formulas or Methods Used

1. Hyperbola definition: $|PF| - |PF'| = \pm 2a$
2. Relation between parameters: $c^2 = a^2 + b^2$
3. Standard form of hyperbola:

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

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Summary of Steps

1. Start with the condition $|PF| - |PF'| = \pm 2a$.
2. Square the equation to eliminate square roots.
3. Expand and isolate terms involving $x$ and $y$.
4. Square again to fully simplify the equation.
5. Rearrange into the standard form of a hyperbola:

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

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Hence, the given condition is proved to form a hyperbola.