Problem 86 - Calculus Early Transcendentals

Question

Consider a hyperbola to be the set of points in a plane whose distances from two fixed points have a constant difference of \(2 a\) or \(-2 a\). Derive the equation of a hyperbola. Assume the two fixed points are on the \(x\) -axis equidistant from the origin.

Step-by-Step Solution

Answer

Question: Given that a set of points in a plane have a constant difference in distances from two fixed points on the x-axis, derive the equation of the hyperbola for these points. Answer: The standard equation of the hyperbola with foci on the x-axis for the given problem is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a\) represents half of the constant difference in distances from the foci, \(b^2 = c^2 - a^2\), and \(c\) is the distance from the origin to each focus.

Step 1: Understanding the Hyperbola and given information

The hyperbola is defined as the set of all points in a plane where the difference between the distances to two fixed points is constant. These two fixed points are called foci, and the constant difference is given as \(2a\) or \(-2a\). In our case, the foci are on the x-axis and are equidistant from the origin. Let's call these foci F1 and F2, and their coordinates \((-c, 0)\) and \((c, 0)\) respectively, where c > a.

Step 2: Analyzing a point on the hyperbola

Let P\((x, y)\) be a point on the hyperbola. According to the definition of the hyperbola, the difference between distances from P to F1 and P to F2 would be \(\pm 2a\). With this information, we can set up an equation using distance formula.

Step 3: Using the distance formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane is given by \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Using this formula, we can find the distances PF1 (distance between Point P and F1) and PF2 (distance between Point P and F2): PF1 = \(\sqrt{(x + c)^2 + y^2}\) PF2 = \(\sqrt{(x - c)^2 + y^2}\)

Step 4: Setting up the difference of distances as \(\pm 2a\)

According to the problem, \(PF2 - PF1 = \pm 2a\). Let's plug in the distance formulas we derived in step 3 and solve for the hyperbola equation. \(\sqrt{(x - c)^2 + y^2} - \sqrt{(x + c)^2 + y^2} = \pm 2a\)

Step 5: Squaring both sides and simplifying

Squaring both sides of the equation to eliminate the square root, we get: \(((x - c)^2 + y^2) - 2\sqrt{(x - c)^2 + y^2}\sqrt{(x + c)^2 + y^2} +((x + c)^2 + y^2)= (2a)^2\) The equation simplifies to: \(- 2\sqrt{(x - c)^2 + y^2}\sqrt{(x + c)^2 + y^2} = 4a^2 - 2c^2 - 2y^2\)

Step 6: Isolating the square root, squaring, and simplifying again

Rearrange the equation and isolate the square root: \(\sqrt{(x - c)^2 + y^2}\sqrt{(x + c)^2 + y^2} = c^2 - a^2 - y^2\) Square both sides again: \(((x - c)^2 + y^2)((x + c)^2 + y^2) = (c^2 - a^2 - y^2)^2\) Expand and simplify the equation to: \(((x^2 - 2cx + c^2) + y^2)((x^2 + 2cx + c^2) + y^2) = c^4 - 2c^2(a^2 + y^2) + (a^2 + y^2)^2\) Simplifying further: \(x^4 + 2(c^2 + y^2)x^2 - c^2(a^2 + y^2) = 0\)

Step 7: Dividing through by \(c^2(a^2 + y^2)\) and simplifying

Divide the equation by \(c^2(a^2 + y^2)\) and obtain the standard form equation of the hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) where \(b^2 = c^2 - a^2\). This is the standard equation of the hyperbola with foci on the x-axis.