Write the equation for the conic section of the cone is centered at (0,0), the major axis is 10 units and the minor axis is 6 units. (Ellipse)
Write the equation for the conic section if the vertex is (-6,4) and the focus is (-6,7) (Parabola)
Write the equation for the conic section if the vertices are at (0,+/-6) and the foci are at (0,+/-10) (Hyperbola)
TYSM FOR ALL OF HELP! but SERIOUS ANSWERS ONLY PLEASE!
All three of these questions fall into the category of basic equations of the particular conic
In your text or class notes, you should have these equations listed along with their particular attributes.
Here is a webpage that summarizes the topic nicely
https://www.varsitytutors.com/hotmath/hotmath_help/topics/conic-sections-and-standard-forms-of-equations
I will do the last one:
Write the equation for the conic section if the vertices are at (0,+/-6) and the foci are at (0,+/-10) (Hyperbola)
We can see that the vertices and foci lie on the y-axis
so we have the form
x^2/a^2 - y^2/b^2 = -1
in a hyperbola: a^2 + b^2 = c^2
for yours:
6^2 + b^2 = 10^2
b^2 = 64
equation:
x^2 / 36 - y^2 / 64 = -1
To write the equations for these conic sections, you can utilize the standard forms for each type of conic section.
1. Equation for an Ellipse:
The standard form of the equation for an ellipse, centered at the origin (0,0), with the major axis of 2a units and the minor axis of 2b units, is:
x^2/a^2 + y^2/b^2 = 1
Given that the major axis is 10 units and the minor axis is 6 units, the equation for the ellipse is:
x^2/5^2 + y^2/3^2 = 1
x^2/25 + y^2/9 = 1
2. Equation for a Parabola:
The standard form of the equation for a parabola, with the vertex at (h,k) and the focus at (h,k+p), is:
(y-k)^2 = 4p(x-h)
Given that the vertex is (-6,4) and the focus is (-6,7), we can determine the value of p, which represents the distance between the vertex and the focus. The distance between the vertex and focus is the same as the distance from the vertex to the directrix. In this case, the directrix would be a horizontal line with y-coordinate k-p. Hence, the directrix is the line y = 1.
Using the distance formula, we can find the value of p:
sqrt((-6 - (-6))^2 + (7 - 4)^2) = sqrt(0^2 + 3^2) = 3.
The equation for the parabola is then:
(y-4)^2 = 4(3)(x-(-6))
(y-4)^2 = 12(x+6)
3. Equation for a Hyperbola:
The standard form of the equation for a hyperbola, centered at the origin (0,0), with the vertices at (0,±a) and the foci at (0,±c), is:
x^2/a^2 - y^2/b^2 = 1
Given that the vertices are at (0,±6) and the foci are at (0,±10), we can determine the values of a and c. The distance between the center and each vertex is a, and the distance between the center and each focus is c.
In this case, a = 6 and c = 10.
The equation for the hyperbola is then:
x^2/6^2 - y^2/√(6^2 + 10^2)^2 = 1
x^2/36 - y^2/196 = 1
(Note: We simplified the expression √(6^2 + 10^2) to 196 for ease of calculation)
Please let me know if you have any further questions!