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Find the area of triangle in parabola #x^2=8y#?

Let #A(x_1, y_1), B(x_2,y_2),C(x_3,y_3)# lie on the parabola #x^2=8y#.
Show that the area of #DeltaABC# is
#1/16(x_1-x_2)(x_2-x_3)(x_3-x_1)#

Answer 1

Sadie Allen

See below.

The area of a triangle given by it's vertices #(x_1,y_1),(x_2,y_2),(x_3,y_3)# is

#A=1/2abs(det((1,1,1),(x_1,x_2,x_3),(y_1,y_2,y_3)))#

now substituting #y_k=x_k^2/8# we have

#A = 1/2det((1,1,1),(x_1,x_2,x_3),(x_1^2/8,x_2^2/8,x_3^2/8))#

#A = 1/(2cdot 8)det((1,1,1),(x_1,x_2,x_3),(x_1^2,x_2^2,x_3^2))#

but

#det((1,1,1),(x_1,x_2,x_3),(x_1^2,x_2^2,x_3^2))=(x_1-x_2)(x_2-x_3)(x_3-x_1)#

#A = 1/16 abs((x_1-x_2)(x_2-x_3)(x_3-x_1))#

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Answer 2

Avery Allen

To find the area of a triangle formed by a parabola, (x^2 = 8y), you'll need to specify the points of intersection between the parabola and the lines forming the triangle. Once you have these points, you can use the formula for the area of a triangle formed by three points in a coordinate plane, typically known as the Shoelace Formula or the Gauss's Area Formula.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Find the area of triangle in parabola #x^2=8y#?

Let #A(x_1, y_1), B(x_2,y_2),C(x_3,y_3)# lie on the parabola #x^2=8y#. Show that the area of #DeltaABC# is #1/16(x_1-x_2)(x_2-x_3)(x_3-x_1)#

Let #A(x_1, y_1), B(x_2,y_2),C(x_3,y_3)# lie on the parabola #x^2=8y#.
Show that the area of #DeltaABC# is
#1/16(x_1-x_2)(x_2-x_3)(x_3-x_1)#