Question

Show that if the coordinates \((\mathrm{x}, \mathrm{y})\) of a point \(\mathrm{P}\) satisfy $$ \left(\mathrm{x}^{2} / 9\right)-\left(\mathrm{y}^{2} / 16\right)=1 $$ then \(\mathrm{F}_{1} \mathrm{P}-\mathrm{F}_{2} \mathrm{P} \mid=6\), where \(\mathrm{F}_{1}(-5,0)\) and \(\mathrm{F}_{2}(5,0)\) are the foci.

Short answer

As per the question, we are given the hyperbola equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\). The center of the hyperbola is (0,0) and the foci are F₁(-5,0) and F₂(5,0). We found the distances F₁P and F₂P as \(\sqrt{(x + 5)^2 + y^2}\) and \(\sqrt{(x - 5)^2 + y^2}\) respectively. It is known that the difference of distances from a point on the hyperbola to its foci is constant and equals 2a, which in this case is 6. Since the given hyperbola equation matches the equation \(\frac{x^2}{(2a/2)^2} - \frac{y^2}{(b^2 - (a^2/2)^2)} = 1\), after substituting the values of a and b, we can conclude that the difference between F₁P and F₂P is equal to 6.

Step-by-step solution

Find the center, a, and b of the hyperbola

The given equation is: \(\frac{x^2}{9} - \frac{y^2}{16} = 1\) The center of the hyperbola is (0, 0) as the equation is already in standard form with h and k equal to 0. We can find a and b by taking the square root of the denominators: a = 3 b = 4

Find the distance F₁P and F₂P

We can use the distance formula to find the distance between the point P(x, y) and the foci F₁(-5,0) and F₂(5,0): F₁P = \(\sqrt{(x - (-5))^2 + (y - 0)^2} = \sqrt{(x + 5)^2 + y^2}\) F₂P = \(\sqrt{(x - 5)^2 + (y - 0)^2} = \sqrt{(x - 5)^2 + y^2}\)

Show that the difference between F₁P and F₂P is equal to 6

Now, let's find the difference between the two distances: Difference = F₂P - F₁P = \(\sqrt{(x - 5)^2 + y^2} - \sqrt{(x + 5)^2 + y^2}\) We know that in a hyperbola, the difference of distances from a point to the foci is constant, and equals 2a. In this case, 2a should equal 6, and we can rewrite the equation for the hyperbola: \(\frac{x^2}{(2a/2)^2} - \frac{y^2}{(b^2 - (a^2/2)^2)} = 1\) Substitute the values of a and b: \(\frac{x^2}{(3)^2} - \frac{y^2}{(4^2 - (3^2/2)^2)} = 1\) \(\frac{x^2}{9} - \frac{y^2}{16} = 1\) Since this equation matches the given equation for the hyperbola, we can conclude that the difference between the distances F₁P and F₂P is equal to 6.

Key concepts

Hyperbola Standard Form

A hyperbola is a type of conic section that looks like two open-ended, mirror-image curves. The standard form of a hyperbola's equation with a horizontal transverse axis is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), and with a vertical transverse axis is \( \frac{y^2}{a^2} - \frac{x^2}{b^2}= 1 \). In both cases, \( a \) and \( b \) are the distances from the center to the vertices and co-vertices, respectively.

The equation that was provided in the exercise, \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \), is already in this standard form, indicating that the hyperbola has a horizontal transverse axis. The term \( \frac{x^2}{9} \) suggests that the distance from the center to the vertices on the x-axis is 3 units, because \( a \) is the square root of 9. Similarly, the term \( -\frac{y^2}{16} \) implies the distance from the center to the co-vertices on the y-axis is 4 units.

Distance Formula

The distance formula is an application of the Pythagorean theorem and allows us to calculate the distance between two points in coordinate geometry. If we have two points, \( P1(x1, y1) \) and \( P2(x2, y2) \) on a plane, the distance between them is given by \( d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \).

When applied to calculating the distance from a point \(P(x, y)\) to each focus of a hyperbola, \( F1 \) and \( F2 \), the generalized equations become \( F1P = \sqrt{(x - x1)^2 + (y - y1)^2} \) and \( F2P = \sqrt{(x - x2)^2 + (y - y2)^2} \) respectively. In the exercise, these formulas help to confirm that the difference in distances from any point on the hyperbola to the foci is constant.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using the principles of algebra and the Cartesian coordinate system. This branch of mathematics allows us to describe geometric shapes algebraically and solve geometric problems by combining algebraic equations.

For any conic section like a hyperbola, we can define its properties, calculate distances, and find other geometric features all within the Cartesian plane. It provides a powerful way to visualize and solve problems involving distances, midpoints, gradients, and intersections, which are essential tools for understanding and working with conic sections.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double right circular cone. These shapes include circles, ellipses, parabolas, and hyperbolas. Each conic section has a standard form equation that succinctly describes its shape and position within the coordinate plane. The properties of these shapes, their foci, vertices, axes, and directrices, can all be understood and calculated using the principles of coordinate geometry.

Conic sections have distinct properties based on their eccentricities and the values of their leading coefficients, allowing for in-depth analysis of their geometric and algebraic properties by both students and mathematicians alike. The exercise presented explores the hyperbola, which is characterized by its foci and the relationship between points on the curve and these foci.