Question

Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse: \[ \frac{x^2}{36} + \frac{y^2}{16} = 1 \]

Hint:

For an ellipse, the length of the major axis is \( 2a \), the length of the minor axis is \( 2b \), the foci are at \( (\pm c, 0) \), and \( c = \sqrt{a^2 - b^2} \).

Answer 1

Step 1: Recognize the standard form of the ellipse equation.
The given equation is in the standard form of an ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where \( a^2 \) is the square of the length of the semi-major axis and \( b^2 \) is the square of the length of the semi-minor axis.
Step 2: Identify \( a^2 \) and \( b^2 \).}
From the equation \( \frac{x^2}{36} + \frac{y^2}{16} = 1 \), we see that: \[ a^2 = 36, \quad b^2 = 16 \] So, \( a = 6 \) and \( b = 4 \).
Step 3: Find the foci of the ellipse.}
The foci of an ellipse are located at \( ( \pm c, 0) \) along the major axis, where \( c \) is given by: \[ c = \sqrt{a^2 - b^2} \] Substitute the values of \( a^2 \) and \( b^2 \): \[ c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \] So, the foci are at \( (\pm 2\sqrt{5}, 0) \).
Step 4: Find the coordinates of the vertices.}
The vertices are located at \( (\pm a, 0) \) along the major axis. Therefore, the vertices are at \( (6, 0) \) and \( (-6, 0) \).
Step 5: Find the length of the major axis.}
The length of the major axis is \( 2a = 2 \times 6 = 12 \).
Step 6: Find the length of the minor axis.}
The length of the minor axis is \( 2b = 2 \times 4 = 8 \).
Step 7: Find the eccentricity.}
The eccentricity \( e \) of an ellipse is given by: \[ e = \frac{c}{a} \] Substitute the values of \( c \) and \( a \): \[ e = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \]
Step 8: Find the length of the latus rectum.}
The length of the latus rectum \( L \) is given by: \[ L = \frac{2b^2}{a} \] Substitute the values of \( b^2 \) and \( a \): \[ L = \frac{2 \times 16}{6} = \frac{32}{6} = \frac{16}{3} \]