What is the equation of the parabola that has a vertex at # (-1, 7) # and passes through point # (2,-3) #?

1 Answer
A. S. Adikesavan
Oct 4, 2016

If the axis is assumed to be parallel to the x-axis, #(y-7)^2=100/3(x+1)# See explanation for the equation of the family of parabolas, when there is no such assumption.

Explanation:

Let the equation of axis of the parabola with vertex #V(-1, 7)# be

#y-7=m(x+1)#, with m not equal tom 0 nor #oo#..

Then the equation of the tangent at the vertex will be

#y-7=(-1/m)(x+1)#.

Now, the equation of any parabola having V as vertex is

#(y-7-m(x+1))^2= 4a(y-7+(1/m)(x+1))#.

This passes through #(2, -3)#, if

#(-10-3m)^2=4a(3/m-10)#. This gives the relation between the two

parameters a and m as

#9m^3+60m^2+(100+40a)m-12a=0#.

In particular, if the axis is assumed to be parallel to the x-axis, m = 0,

this method can be ignored.

In this case, #y-7=0# is for the axis and x+1 = 0 is for the tangent at

the vertex. and the equation of the parabola becomes

#(y-7)^2=4a(x+1).#

As it passes through (2, -3), a =25/3.

The parabola is given by

#(y-7)^2=100/3(x+1)#

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