Problem 3 - Precalculus

Question

You are given an ellipse and a point P on the ellipse. Find \(F_{1} P\) and \(F_{2} P\), the lengths of the focal radii. $$\left(x^{2} / 15^{2}\right)+\left(y^{2} / 5^{2}\right)=1 ; P(9,4)$$

Step-by-Step Solution

Answer

\(F_1P \approx 6.51\), \(F_2P \approx 23.47\).

Step 1: Identify the Foci of the Ellipse

The equation of the ellipse is \(\frac{x^2}{15^2} + \frac{y^2}{5^2} = 1\). It is centered at the origin with semi-major axis \(a = 15\) and semi-minor axis \(b = 5\). For this horizontal ellipse, the foci are located at \((\pm c, 0)\) where \(c = \sqrt{a^2 - b^2}\). Calculate \(c\).

Step 2: Calculate the Value of c

Using the formula \(c = \sqrt{a^2 - b^2}\), plug in \(a = 15\) and \(b = 5\):\[ c = \sqrt{15^2 - 5^2} = \sqrt{225 - 25} = \sqrt{200} \approx 14.14 \] Thus, the foci are at \((14.14, 0)\) and \((-14.14, 0)\).

Step 3: Compute Distance \(F_1P\)

Using the distance formula, compute the distance from the point \(P(9,4)\) to the focus \(F_1(14.14,0)\): \[ F_1P = \sqrt{(9 - 14.14)^2 + (4 - 0)^2} = \sqrt{(-5.14)^2 + 4^2} \approx \sqrt{26.4196 + 16} \approx \sqrt{42.4196} \approx 6.51 \]

Step 4: Compute Distance \(F_2P\)

Similarly, compute the distance from \(P(9,4)\) to the focus \(F_2(-14.14,0)\): \[ F_2P = \sqrt{(9 + 14.14)^2 + (4 - 0)^2} = \sqrt{23.14^2 + 4^2} \approx \sqrt{535.3296 + 16} \approx \sqrt{551.3296} \approx 23.47 \]

Key Concepts

Foci of an Ellipse

An ellipse is defined by two prominent fixed points called foci. These foci are essential to understanding the shape and properties of the ellipse. In a horizontal ellipse, like the one given in the problem with the equation \( \frac{x^2}{15^2} + \frac{y^2}{5^2} = 1 \), the foci are symmetrically located along the x-axis.
To find the foci, we need to calculate the value of \( c \) where the foci are at positions \((\pm c, 0)\). The formula to determine \( c \) given the semi-major axis \( a \) and the semi-minor axis \( b \) is:
\[ c = \sqrt{a^2 - b^2} \]
For this particular ellipse:

• \( a = 15 \), the semi-major axis
• \( b = 5 \), the semi-minor axis

Plugging in these values, we find that \( c \approx 14.14 \), so the foci of the ellipse are located at \((14.14, 0)\) and \((-14.14, 0)\). These foci influence how stretched out the ellipse appears and play a vital role in defining its overall geometry.

Distance Formula

The distance formula is a powerful tool used to determine the length between two points in a plane. The formula itself is based on the Pythagorean theorem and is given by:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula calculates the straight-line distance between the points \( (x_1, y_1) \) and \( (x_2, y_2) \).
In the context of the ellipse problem, we use this formula to find the distances \( F_1P \) and \( F_2P \), where \( P \) is the point on the ellipse at \( (9, 4) \), and \( F_1 \) and \( F_2 \) are the foci.

• For \( F_1P \), substitute \( F_1(14.14, 0) \) and \( P(9, 4) \) into the formula to get:\[ F_1P = \sqrt{(9 - 14.14)^2 + (4 - 0)^2} \]
• For \( F_2P \), substitute \( F_2(-14.14, 0) \) and \( P(9, 4) \) into the formula to get:\[ F_2P = \sqrt{(9 + 14.14)^2 + (4 - 0)^2} \]

This calculation enables us to find how far a specific point on the ellipse is from each focus, an important aspect of understanding ellipses.

Semi-Major Axis

The semi-major axis is a crucial part of an ellipse's equation and contributes significantly to its shape. In any ellipse, the semi-major axis is the longest radius from the center to the perimeter of the ellipse. In mathematical terms, it's represented by \( a \) in the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
In the specific ellipse from our problem, the semi-major axis \( a \) is along the x-axis, with a length of 15. The quantity \( a \) determines how elongated the ellipse is horizontally.
Understanding the semi-major axis is important in solving problems involving foci and distances since:

• It helps determine the position of the foci via the formula \( c = \sqrt{a^2 - b^2} \).
• It indicates the maximum distance from the center of the ellipse to a point on its edge in its longest direction.

This makes the semi-major axis a key component when examining the characteristics and measurements related to ellipses.