No, that's completely wrong! What happened to the y2 in the original equation? Try completing the square in y again!duki said:Homework Statement
graph the following:
Homework Equations
[tex]x^2-4y^2+2x+8y-7=0[/tex]
The Attempt at a Solution
So far I've gotten to [tex](x+1)^2 = -4(y-2)[/tex]
If that's right, I have p = -1 and v = (-1 , 2) and directrix: y=2. Could someone double check me to see if I'm doing it right?
Thanks
A conic section is a curve that is formed by the intersection of a plane with a cone. There are four types of conic sections: circles, ellipses, parabolas, and hyperbolas.
To find the vertex of a conic section, you can use the formula (-b/2a, -d/2a), where a and b are the coefficients of the x-squared and y-squared terms, and d is the constant term. For example, in the equation x^2-4y^2+2x+8y-7=0, the vertex would be (-1, 1).
The directrix of a conic section is a line that is perpendicular to the axis of symmetry and located at a distance from the vertex equal to the distance from the focus to the vertex. In other words, it is the line that the curve reflects off of. For the equation x^2-4y^2+2x+8y-7=0, the directrix would be y=1.
The focus of a conic section is a point that is located on the axis of symmetry and at a distance from the vertex equal to the distance from the vertex to the directrix. In the equation x^2-4y^2+2x+8y-7=0, the focus would be (1, 1).
Yes, you can graph a conic section without knowing the equation by using a geometric construction. For example, to graph an ellipse, you can use two pins and a string to create the curve. To graph a parabola, you can use a piece of paper and a pencil to create the reflective property of the curve. However, knowing the equation can help you find specific points on the curve, such as the vertex, directrix, and focus.