Solving the Ellipse Point Distance Problem

The problem asks for the sum of the distances from any point P(x, y) on the ellipse given by the equation \(25x^2 + 16y^2 = 400\) to two specific points, Q(0, 3) and R(0, -3). This involves understanding the properties of an ellipse, particularly the role of its foci.

Standard Form of the Ellipse Equation

First, let's convert the given equation of the ellipse into its standard form to identify key parameters.

The given equation is: \(25x^2 + 16y^2 = 400\)

To get the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we divide the entire equation by 400:

\[ \frac{25x^2}{400} + \frac{16y^2}{400} = \frac{400}{400} \]

\[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \]

Comparing this to the standard form, we have:

• \(a^2 = 16 \implies a = 4\)
• \(b^2 = 25 \implies b = 5\)

Since \(b > a\), the major axis of the ellipse is along the y-axis, and the minor axis is along the x-axis. The center of the ellipse is at the origin (0, 0).

Identifying the Foci of the Ellipse

The foci of an ellipse lie on the major axis. For an ellipse with the major axis along the y-axis, the foci are located at \((0, \pm c)\), where \(c\) is the distance from the center to each focus. The relationship between \(a\), \(b\), and \(c\) is given by \(c^2 = b^2 - a^2\) for \(b > a\).

Using the values we found:

\[ c^2 = 5^2 - 4^2 \]

\[ c^2 = 25 - 16 \]

\[ c^2 = 9 \]

\[ c = \sqrt{9} = 3 \]

So, the coordinates of the foci are \((0, \pm 3)\). These are F\(_1\)(0, 3) and F\(_2\)(0, -3).

Relating Given Points Q and R to Foci

The given points are Q(0, 3) and R(0, -3).

Comparing these coordinates to the foci we just calculated, we see that the points Q and R are precisely the foci of the given ellipse.

• Q(0, 3) is one focus.
• R(0, -3) is the other focus.

Ellipse Property: Sum of Distances to Foci

A fundamental property of any point P on an ellipse is that the sum of the distances from P to the two foci is a constant value. This constant value is equal to the length of the major axis.

Let P(x, y) be any point on the ellipse. Let F\(_1\) and F\(_2\) be the foci. Then, PF\(_1\) + PF\(_2\) = Length of the major axis.

Calculating the Sum of Distances PQ + PR

Since Q and R are the foci of the ellipse, the sum of the distances PQ + PR for any point P on the ellipse is equal to the length of the major axis.

The major axis is along the y-axis, and its length is \(2b\).

Length of major axis = \(2 \times b = 2 \times 5 = 10\).

Therefore, for any point P(x, y) on the ellipse \(25x^2 + 16y^2 = 400\), the sum of the distances PQ + PR is equal to 10.

In summary, the steps are:

1. Convert the ellipse equation to standard form to identify \(a\) and \(b\).
2. Determine the orientation of the major axis and calculate \(c\).
3. Find the coordinates of the foci.
4. Recognize that the given points Q and R are the foci.
5. Apply the property that the sum of distances from any point on the ellipse to its foci is equal to the length of the major axis.
6. Calculate the length of the major axis.

The calculation confirms that PQ + PR = 10.

Revision Table: Key Ellipse Concepts

Additional Information on Ellipse Properties

The ellipse is one of the conic sections, formed by the intersection of a plane and a cone. Its unique property regarding the sum of distances to the foci is often used as its definition. This property has practical applications, such as in optics (elliptical reflectors) and in astronomy (planetary orbits are elliptical with the sun at one focus).

The eccentricity (\(e\)) of an ellipse is another important parameter, defined as \(e = c / (\text{semi-major axis})\). It measures how elongated the ellipse is. For an ellipse, \(0 < e < 1\). An eccentricity close to 0 means the ellipse is nearly circular, while an eccentricity close to 1 means it is very elongated.

For the given ellipse \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), the semi-major axis is \(b=5\) and \(c=3\). The eccentricity is \(e = c/b = 3/5 = 0.6\).