Question

Find the equation of the set of points \(P\) satisfying the given conditions. The difference of the distances of \(P\) from \((0, \pm 6)\) is 10 .

Short answer

The equation is \(\frac{y^2}{25} - \frac{x^2}{11} = 1\).

Step-by-step solution

Identify Key Information

We need to find the equation of the locus of points \(P(x, y)\) such that the difference between the distances from \(P\) to the points \((0, 6)\) and \((0, -6)\) is 10. This is the geometric definition of a hyperbola.

Use the Hyperbola Definition

The standard form of a hyperbola given the difference of distances from two fixed points (foci) \( (0, c) \) and \( (0, -c) \) is constant is: \(|d_1 - d_2| = 2a\), where \(d_1\) and \(d_2\) are the distances to the foci. In our problem, \(2a = 10\), so \(a = 5\). The foci are \((0, 6)\) and \((0, -6)\), giving \(c = 6\). The difference of distances being constant implies \(c^2 - a^2 = b^2\).

Calculate \(b\)

Given \(c = 6\) and \(a = 5\), use the formula \(c^2 - a^2 = b^2\) to find \(b\). We get \(b^2 = 6^2 - 5^2 = 36 - 25 = 11\). So, \(b = \sqrt{11}\).

Formulate the Hyperbola Equation

The equation of the hyperbola with center at the origin and transverse axis along the y-axis is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Substituting \(a = 5\) and \(b = \sqrt{11}\), we get \(\frac{y^2}{25} - \frac{x^2}{11} = 1\).

Final Equation

The final equation representing the locus of points \(P\) such that the difference of distances to the points \((0, 6)\) and \((0, -6)\) is 10 is \(\frac{y^2}{25} - \frac{x^2}{11} = 1\).

Key concepts

Locus of Points

In geometry, the term "locus" refers to a set of points that satisfy a particular condition. When dealing with hyperbola equations, the locus of points is critical as it defines the path or shape that adheres to specific geometric principles. A hyperbola is essentially the locus of all points for which the absolute difference of the distances to two fixed focal points is constant.

For instance, imagine P(x, y) as a point in space. If you were to track this point such that the difference between its distances to two fixed points

• (0, 6)
• (0, -6)

is always a constant number, like 10 in this particular problem, you would find that P traces out the shape of a hyperbola.

Understanding loci is vital in conquering mysteries around shapes formed by such conditions. It's a bridge connecting conditions to tangible geometric forms.

Hyperbola Foci

The foci of a hyperbola are integral in defining the hyperbola's shape and properties. By definition, a hyperbola has two fixed points known as foci. The role these points play is key: they determine all points that belong to the hyperbola by ensuring the constant difference in distances.

In our exercise, the foci are given as

• (0, 6)
• (0, -6)

These points lie on the y-axis, showing us that our hyperbola is oriented vertically. The distance from the center (origin) to each focus, denoted by c, is 6 in this scenario.

Knowing the foci helps in forming the hyperbola equation as it provides the value of c, which also helps compute other vital parameters like a (semi-major axis) and b (semi-minor axis) through mathematical relations.

Conic Sections

Conic sections are curves obtained by intersecting a double cone (a cone with two identical nappes) with a plane. These intersections give rise to different types of curves including circles, ellipses, parabolas, and hyperbolas. In context, a hyperbola emerges when the plane cuts both nappes of the cone at an angle to the axis that is steeper than the cone's side.

Understanding hyperbolas as part of conic sections allows us to appreciate their properties and the relationships between different geometric systems. It unifies various geometric shapes under a single framework, linking the circle, ellipse, and hyperbola through variations in intersection angles.

This fundamental idea elevates our comprehension in geometry, extending beyond the simple visualization of hyperbolas. It roots them into the rich, interconnected world of conic sections.

Geometry of Hyperbolas

The geometry of hyperbolas is fascinating due to their unique structure characterized by two distinct curves. These curves, or branches, open in opposite directions. In this exercise, since the foci are positioned vertically, the branches of the hyperbola open along the y-axis.

The equation \(\frac{y^2}{25} - \frac{x^2}{11} = 1\)represents the geometry of our hyperbola. Here, we see that

• \(a = 5\) is half the length of the transverse axis, indicating vertical stretch.
• \(b = \sqrt{11}\) serves as half the conjugate axis's length, guiding the hyperbola's width.

Hyperbolas are also known for their asymptotes, which are straight lines that the branches approach but never actually intersect. These lines give the hyperbola its ever-opening shape and are calculated using the terms a and b. Capturing these geometry principles uncovers the elegance and complexity of hyperbolas, revealing their practical uses and roles in various mathematical problems.