Find the equation, Centre, vertices, co-vertices, foci, equations of assymptotes and eccentricity of the hyperbola traced by the points that moves so that the differences of its distance from (-4,4) and (-4,6) is 6.
google is your friend. Try this to get going:
https://people.richland.edu/james/lecture/m116/conics/hypdef.html
I think the question have mistake.
I tried my best but the graph is not like hyperbola.
The value of a is greater than c therefore, b is undefined
I agree. Since the distance between the foci is only 2, there is no way that the condition can be met.
To find the equation, center, vertices, co-vertices, foci, equations of asymptotes, and eccentricity of the hyperbola described by the given condition, we can follow these steps:
Step 1: Understanding the Hyperbola Equation
The standard form of a hyperbola equation with its center at (h, k) is:
((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1, for a horizontal hyperbola
or
((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1, for a vertical hyperbola
Step 2: Finding the Center
From the given information, the differences of the distances from (-4, 4) and (-4, 6) is 6. This implies that the center of the hyperbola will be at (-4, 5) since the y-coordinate value that splits the difference between 4 and 6 is 5.
So, the center of the hyperbola is (h, k) = (-4, 5).
Step 3: Finding the Distance from the Center to the foci and Vertices
By definition, the difference between the distances from any point on a hyperbola to its two foci is constant. In this case, the difference of distances is given as 6.
From the center (h, k) = (-4, 5), the distance to one of the foci, denoted as "c," can be calculated by the formula: c = sqrt(a^2 + b^2), where a and b are the semi-major and semi-minor axes.
Given that the distance between (-4, 4) and (-4, 6) is 6, we can conclude that "c" is exactly equal to 3.
Step 4: Finding the Value of a
To calculate the value of "a," we know that one semi-major axis (a) extends from the center to one of the vertices. Since the vertical distance between the center and one of the vertices is 3, the value of "a" will be 3.
Step 5: Finding the Value of b
To calculate the value of "b," we know that one semi-minor axis (b) extends from the center to one of the co-vertices. The distance between the center and one of the co-vertices can be found using the Pythagorean theorem:
b^2 = c^2 - a^2
b^2 = 3^2 - 3^2
b^2 = 9 - 9
b^2 = 0
From this calculation, we find that "b" is equal to 0.
Step 6: Obtaining the Equation of the Hyperbola
Using the values of a, b, and the center (h, k), we can now write the equation of the hyperbola. Since the value of b is 0, we have a vertical hyperbola with the equation:
((y - 5)^2 / 3^2) - ((x + 4)^2 / 0^2) = 1
Simplifying further, we get:
(y - 5)^2 / 9 = 1
Step 7: Finding the Asymptotes and Eccentricity
For a vertical hyperbola, the equations of the asymptotes can be written as follows:
y - k = +/- (a / b)(x - h)
Substituting the values of a, b, h, and k, we can calculate the two equations of the asymptotes:
y - 5 = +/- (3 / 0)(x + 4)
As b is equal to 0, the denominator becomes zero, resulting in a vertical line at x = -4.
Lastly, the eccentricity of the hyperbola can be calculated using the formula:
eccentricity = c / a
Substituting the values of c and a, we get:
eccentricity = 3 / 3
eccentricity = 1
In summary, the equation of the hyperbola is ((y - 5)^2 / 9) - ((x + 4)^2 / 0) = 1, with its center at (-4, 5), no real value for b (semi-minor axis), foci at (-4, 2) and (-4, 8), eccentricity of 1, and the equation of the asymptotes as x = -4.