How do you find the equation of a parabola when given two points (2,10) and (4,10)?

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Answer 1 · 2015-07-26T00:09:05

There's not enough information here to uniquely determine the parabola, but enough to find:

#y = ax^2-6ax+(8a+10)# for some constant #a in RR#

Since the #y# coordinate of both points is the same, the axis must be midway between the two points at #x=3#.

So the equation of the parabola takes the form:

#y = a(x-3)^2+k#

If we substitute #x = 4# and #y = 10# into this equation we get:

#10 = a(4-3)^2 + k = a+k#

So #k = 10-a# and we can write the equation as:

#y = a(x-3)^2+(10-a)#

#=ax^2-6ax+9a+10-a#

#=ax^2-6ax+(8a+10)#

The constant #a# can be given any value in #RR#, except that when #a=0# the 'parabola' is the straight line #y=10#.

Here are the parabolas for #a=+-1# and #a=+-2#

graph{(y-x^2+6x-18)(y+x^2-6x-2)(y-2x^2+12x-26)(y+2x^2-12x+6)=0 [-2.585, 7.415, 7.22, 12.22]}