No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola, and every point on the parabola satisfies the equation.
If the equation of the parabola isy = ax^2 + bx + c then the roots are [-b +/- sqrt(b^2-4ac)]/(2a)
Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
You don't really solve the equation. You use it. Having said that, see the Wikipedia article, which has an adequate discussion of the equation and shows it in a few forms.
First you need more details about the parabola. Then - if the parabola opens upward - you can assume that the lowest point of the triangle is at the vertex; write an equation for each of the lines in the equilateral triangle. These lines will slope upwards (or downwards) at an angle of 60°; you must convert that to a slope (using the tangent function). Once you have the equation of the lines and the parabola, solve them simultaneously to check at what points they cross. Finally you can use the Pythagorean Theorem to calculate the length.
A parabola is a graph of a 2nd degree polynomial function. Two graph a parabola, you must factor the polynomial equation and solve for the roots and the vertex. If factoring doesn't work, use the quadratic equation.
No you can't. There is no unique solution for 'x' and 'y'. The equation describes a parabola, and every point on the parabola satisfies the equation.
If the equation of the parabola isy = ax^2 + bx + c then the roots are [-b +/- sqrt(b^2-4ac)]/(2a)
If you know the equation, you just plug in x = 0 and solve.
The focus of a parabola is a fixed point that lies on the axis of the parabola "p" units from the vertex. It can be found by the parabola equations in standard form: (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) depending on the shape of the parabola. The vertex is defined by (h,k). Solve for p and count that many units from the vertex in the direction away from the directrix. (your focus should be inside the curve of your parabola)
Set y = 0 and solve for x, with a parabola you should get one, two, or no x-axis crossings, it depends on the equation and the location on the x-y axis of the parabola.
Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
The formula for the Latus rectum is simply 2L = 4a with a stands for the distance of the focus from the vertex of the parabola. Given a, you can simply solve for the length of the latus rectum by using this formula.. L = 2a
y=b+x+x^2 This is a quadratic equation. The graph is a parabola. The quadratic equation formula or factoring can be used to solve this.
Y=a(x-h)+k is the vertex formula. Since the vertex is at (-2,-3) this parabola has the equation: y=a(x+2)^2-3 We can plug in x=-1 but we really need to know a, to solve for y. ( we can solve it, but we will have an a in the solution)
You don't really solve the equation. You use it. Having said that, see the Wikipedia article, which has an adequate discussion of the equation and shows it in a few forms.
First you need more details about the parabola. Then - if the parabola opens upward - you can assume that the lowest point of the triangle is at the vertex; write an equation for each of the lines in the equilateral triangle. These lines will slope upwards (or downwards) at an angle of 60°; you must convert that to a slope (using the tangent function). Once you have the equation of the lines and the parabola, solve them simultaneously to check at what points they cross. Finally you can use the Pythagorean Theorem to calculate the length.