Question 30638: write the equation of a parabola that opens downward with an axis of symmetry of x=5/2. include a graph of the parabola
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! Since only the axis of symmetry x =5/2 ( a line parallel to the y-axis at a distance 5/2 units to the right of y-axis)is given, you can generate a whole family of parabolas with a = the distance of the focus from the vertex fixed and since the parabola is looking downwards the focus for every suh parabola should be always at a distance = a below the vertex.
Both the vertex A and the focus S lie on x = 5/2
Therefore both have x- cordinate equal to 5/2.
And if A has y-co-ordinate = k then since S is at a distance a units below A, therefore S will have y-co-ordinate = k-a
Here k is not fixed. Corresponding to every value of k we get a parabola looking downward with its vertexA and focus S on x = 5/2 and its equation given by (x-5/2)^2 = -4a(y-k)----(I) (with a fixed)
(the minus sign outside on the RHS is because the parabola is looking downwards.
Note: supposing we keep k fixed and allow S to travel on the axis of symmetry
then the equation to every such member though will look the same
(x-5/2)^2 = -4a(y-k)----(I) (with k fixed)
Request: As I have no facility to draw a graph inside this answer box please convert my explanations into graph yourself
|
|
|
