What is the eccentricity of the hyperbola 9x 2 16y 2 144?

⇒ e = 5/4. Hence (B) is the correct answer.

How to find eccentricity of an ellipse?

The formula to determine the eccentricity of an ellipse is the distance between foci divided by the length of the major axis.

What kind of conic section is the 9×2 16y 144?

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

What is the eccentricity of ellipse 9x 2 5y 2 18x 20y 16 0?

Hence, eccentricity, e=⅔.

Which of the following is the foci of the hyperbola 9x 2 16y 2 144?

This becomes a standard form of the hyperbola with transverse axis \[a = 4\] and conjugate axis \[b = 3\]. The foci of the hyperbola are of the form \[\left( ae, 0 \right)\] and \[\left( – ae, 0 \right)\]. Therefore, the foci are \[\left( 5, 0 \right)\] and \[\left( – 5, 0 \right)\] .

What is the eccentricity of hyperbola?

The eccentricity of an ellipse which is not a circle is greater than zero but less than 1. The eccentricity of a parabola is 1. The eccentricity of a hyperbola is greater than 1.

How do you find eccentricity?

Find the eccentricity of an ellipse. This is given as e = (1-b^2/a^2)^(1/2). Note that an ellipse with major and minor axes of equal length has an eccentricity of 0 and is therefore a circle.

Which of the following is the eccentricity of ellipse?

4. Which of the following is the eccentricity for an ellipse? Explanation: The eccentricity for ellipse is always less than 1. The eccentricity is always 1 for any parabola.

How do you factor 9x 2 16y 2?

1. Using:a2−b2=(a+b)(a−b)
2. 9×2−16y2=(3x)2−(4y)2=(3x+4y)(3x−4y)
3. Hence, 9×2−16y2=(3x+4y)(3x−4y)

What are the foci of the hyperbola with equation 16y2 9×2 144?

The first focus of a hyperbola can be found by adding c c to k k . Substitute the known values of h h , c c , and k k into the formula and simplify. The second focus of a hyperbola can be found by subtracting c c from k k . Substitute the known values of h h , c c , and k k into the formula and simplify.

What is the eccentricity of conjugate hyperbola?

The eccentricity of the conjugate hyperbola is given by a2 = b2(e2 – 1).

Which of the following is eccentricity of ellipse?

How to find the distance from center to focus of ellipse?

Find the distance from the center to a focus of the ellipse by using the following formula. Substitute the values of a a and b b in the formula. Simplify. Tap for more steps… Raise 4 4 to the power of 2 2. Raise 3 3 to the power of 2 2. Multiply − 1 – 1 by 9 9. Subtract 9 9 from 16 16. Find the vertices. Tap for more steps…

What are the important values for graphing an ellipse?

Raise 3 3 to the power of 2 2. Multiply − 1 – 1 by 9 9. Subtract 9 9 from 16 16. These values represent the important values for graphing and analyzing an ellipse.

How do you find the standard form of an ellipse?

Find the standard form of the ellipse. Tap for more steps… Divide each term by 144 144 to make the right side equal to one. Simplify each term in the equation in order to set the right side equal to 1 1. The standard form of an ellipse or hyperbola requires the right side of the equation be 1 1. This is the form of an ellipse.

How do you find the first vertex of an ellipse?

Raise 4 4 to the power of 2 2. Raise 3 3 to the power of 2 2. Multiply − 1 – 1 by 9 9. Subtract 9 9 from 16 16. Find the vertices. Tap for more steps… The first vertex of an ellipse can be found by adding a a to h h. Substitute the known values of h h, a a, and k k into the formula. Simplify.