PARABOLA

EQUATION FOR PARABOLA IS : y^2=4x

Kristin!

I recommend sketching this on a piece of graph paper. Draw the parabola with its vertex at the origin and the legs upward. The isosceles triangle will be inverted with the base at the top and its vertex at the origin.

Remember that an isosceles triangle has equal legs and a right isosceles triangle has a right angle. This right angle would be the one at the vertex.

Think about lines and the equation of a line for a second. A line has a formulae of y=mx+b, and two of the legs of the isosceles triangle are described by lines through the origin, so the b intercept is zero and the legs lines become of the form y=mx.

I'll call these triangle leg lines y=m1x and y=m2x.

Now consider that these two lines are perpindicular to each other. for this to occur then m1*m2=-1.

Now go back to the drawing and consider some properties of the triangle, that is the sum of angles is 180deg and one angle is 90des and the triangle is isosceles. The other two angles are then 45 deg. By simple geometry you can see that the angle of the triangle lines are when the rise is equal to the run, or the slope of the right leg of the triangle (m1) is 1, and since m1*m2=-1 then the slope of the left leg is -1.

So the line formulae of the triangle is y=x and y=-x.

Now combine these lines with the parabolas y^2=4x to fine the intersections

first y=x, then y^2=4y or y^2-4y=y(y-4)=0
and solutions for y are y=0 and y=4

then y=-x, then y^2=-4y or y^2+4y=y(y+4)=0
and solutions are y=0 and y=-4.

plugging these solutions into the line legs y=x and y=-x and the vertices of the triangle become (0,0)(4,4), & (-4,4)

Now look at the triangle. It has a height of 4 (the distance from the origin to the base) and a base of 8 [4-(-4)]. Using the fermulae for the area of a triangle (1/2bh) this becomes 1/2*8*4=16

Rap

PARABOLA
Hello!

Thank you Rap for spending your time on this(my) problem.
I` promise wont write at least a month.. (OK!! MAYBE TWO WEEKS..OR??)

Anyway thanks,
Kristin!