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We can draw 3 normals to a parabola from a given point as the equation of normal in parametric form is a cubic equation.
At any point on the y-axis, the x-coordinate is zero. In the equation of the parabola, set x=0. Tidy it up, and you have " Y = the y-intercept ".
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.
It is the parabola such that the coordinates of each point on it satisfies the given equation.
If you want to sketch graphs you have to observe the parabola first then find the vertex afterwards you connect them and you've arrived at your answer. In order to write equations for parabolas it has to have x square in it. The standard equation for a parabola is (y - k)2 = 4a(x - h) where h and k are the x- and y-coordinates of the vertex of the parabola and 'a' is a non zero real number. This website at the related link should help, for the equation at least. A parabola is a basic U shaped graph that meets at one point called a vertex. The equation for Andy parabola must have a number being squared such as x2.
It depends on where points h and k are, in which parabola. Since you have chosen not to share that information, there cannot be any sensible answer to this question.
We can draw 3 normals to a parabola from a given point as the equation of normal in parametric form is a cubic equation.
This is called the 'standard form' for the equation of a parabola:y =a (x-h)2+vDepending on whether the constant a is positive or negative, the parabola will open up or down.
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At any point on the y-axis, the x-coordinate is zero. In the equation of the parabola, set x=0. Tidy it up, and you have " Y = the y-intercept ".
It is the parabola such that the coordinates of each point on it satisfies the given 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.
Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2
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There are several ways of defining a parabola. Here are some:Given a straight line and a point not on that line, a parabola is the locus of all points that are equidistant from that point (the focus) and the line (directrix).A parabola is the intersection of the surface of a right circular cone and a plane parallel to a generating line of that surface.A parabola is the graph of a quadratic equation.
The coordinates will be at the point of the turn the parabola which is its vertex.