Right Triangle Inscribed in Parabola

T

turophile

Junior Member
Joined
May 22, 2010
Messages
94
A right triangle of area 20 is inscribed in the parabola y = ¼x^2, with the vertex of the right angle at the origin. What are the other two vertices of the triangle?

Here's what I have done so far: Let C be the length of the hypotenuse, and A and B be the lengths of the sides meeting at the origin. Let(x1, y1) and (0, 0) be the endpoints of side A, and let (x2, y2) be the endpoints of side B. Let L1 be the line that includes side A, and let L2 be the line that includes L2. Then the following are true:

y1 = (1/4)(x1)^2
y2 = (1/4)(x2)^2
A = sqrt((x1 - 0)^2 + (y1 - 0)^2) = sqrt((x1)^2 + (y1)^2)
B = sqrt((x2 - 0)^2 + (y2 - 0)^2) = sqrt((x2)^2 + (y2)^2)
AB/2 = 20 ? AB = 40
the slope of L1 is y1/x1
the slope of L2 is y2/x2 = - x1/y1

I'm not sure where to go from here. Any hints? Thanks.
 
turophile said:
A right triangle of area 20 is inscribed in the parabola y = ¼x^2, with the vertex of the right angle at the origin. What are the other two vertices of the triangle?

Here's what I have done so far: Let C be the length of the hypotenuse, and A and B be the lengths of the sides meeting at the origin. Let(x1, y1) and (0, 0) be the endpoints of side A, and let (x2, y2) be the endpoints of side B. Let L1 be the line that includes side A, and let L2 be the line that includes L2. Then the following are true:

y1 = (1/4)(x1)^2
y2 = (1/4)(x2)^2
A = sqrt((x1 - 0)^2 + (y1 - 0)^2) = sqrt((x1)^2 + (y1)^2)
B = sqrt((x2 - 0)^2 + (y2 - 0)^2) = sqrt((x2)^2 + (y2)^2)
AB/2 = 20 ? AB = 40
the slope of L1 is y1/x1
the slope of L2 is y2/x2 = - x1/y1

I'm not sure where to go from here. Any hints? Thanks.
Click to expand...

If I were to do this problem - I'll assume two points

A (x[sub:dcqrfqbt]A[/sub:dcqrfqbt] , x[sub:dcqrfqbt]A[/sub:dcqrfqbt][sup:dcqrfqbt]2[/sup:dcqrfqbt]/4)

and point B (x[sub:dcqrfqbt]B[/sub:dcqrfqbt] , x[sub:dcqrfqbt]B[/sub:dcqrfqbt][sup:dcqrfqbt]2[/sup:dcqrfqbt]/4)

where AO and OB are two perpendicular sides and AB is the hypotenuese.

AO[sup:dcqrfqbt]2[/sup:dcqrfqbt] = (x[sub:dcqrfqbt]A[/sub:dcqrfqbt])[sup:dcqrfqbt]2[/sup:dcqrfqbt] + (x[sub:dcqrfqbt]A[/sub:dcqrfqbt][sup:dcqrfqbt]2[/sup:dcqrfqbt]/4)[sup:dcqrfqbt]2[/sup:dcqrfqbt] = (x[sub:dcqrfqbt]A[/sub:dcqrfqbt])[sup:dcqrfqbt]2[/sup:dcqrfqbt] + (x[sub:dcqrfqbt]A[/sub:dcqrfqbt])[sup:dcqrfqbt]4[/sup:dcqrfqbt]/16

OB[sup:dcqrfqbt]2[/sup:dcqrfqbt] = (x[sub:dcqrfqbt]B[/sub:dcqrfqbt])[sup:dcqrfqbt]2[/sup:dcqrfqbt] + (x[sub:dcqrfqbt]B[/sub:dcqrfqbt][sup:dcqrfqbt]2[/sup:dcqrfqbt]/4)[sup:dcqrfqbt]2[/sup:dcqrfqbt]= (x[sub:dcqrfqbt]B[/sub:dcqrfqbt])[sup:dcqrfqbt]2[/sup:dcqrfqbt] + (x[sub:dcqrfqbt]B[/sub:dcqrfqbt])[sup:dcqrfqbt]4[/sup:dcqrfqbt]/16

AB[sup:dcqrfqbt]2[/sup:dcqrfqbt] = (x[sub:dcqrfqbt]B[/sub:dcqrfqbt] - x[sub:dcqrfqbt]A[/sub:dcqrfqbt])[sup:dcqrfqbt]2[/sup:dcqrfqbt] + (x[sub:dcqrfqbt]B[/sub:dcqrfqbt][sup:dcqrfqbt]2[/sup:dcqrfqbt]/4 - x[sub:dcqrfqbt]B[/sub:dcqrfqbt][sup:dcqrfqbt]2[/sup:dcqrfqbt]/4)[sup:dcqrfqbt]2[/sup:dcqrfqbt]

Now continue
.
 
AB[sup:2h9jmxfk]2[/sup:2h9jmxfk] = (x[sub:2h9jmxfk]B[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] - 2x[sub:2h9jmxfk]A[/sub:2h9jmxfk]x[sub:2h9jmxfk]B[/sub:2h9jmxfk] + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + (x[sub:2h9jmxfk]B[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]/16 - 2(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk](x[sub:2h9jmxfk]B[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk]/4 + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]/16

Then, by the Pythagorean Theorem, AB[sup:2h9jmxfk]2[/sup:2h9jmxfk] = AO[sup:2h9jmxfk]2[/sup:2h9jmxfk] + OB[sup:2h9jmxfk]2[/sup:2h9jmxfk] ? - 2x[sub:2h9jmxfk]A[/sub:2h9jmxfk]x[sub:2h9jmxfk]B[/sub:2h9jmxfk] - 2(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk](x[sub:2h9jmxfk]B[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk]/4 = 0 ? 1 + x[sub:2h9jmxfk]A[/sub:2h9jmxfk]x[sub:2h9jmxfk]B[/sub:2h9jmxfk]/4 = 0 ? x[sub:2h9jmxfk]A[/sub:2h9jmxfk]x[sub:2h9jmxfk]B[/sub:2h9jmxfk] = - 4 ? x[sub:2h9jmxfk]B[/sub:2h9jmxfk] = (- 4/x[sub:2h9jmxfk]A[/sub:2h9jmxfk]),
since x[sub:2h9jmxfk]A[/sub:2h9jmxfk] ? 0.

Substituting for x[sub:2h9jmxfk]B[/sub:2h9jmxfk], OB[sup:2h9jmxfk]2[/sup:2h9jmxfk] = (- 4/x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + (- 4/x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]/16 = 16/(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16/(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]

Since the triangle has area = 20, (AO)(OB)/2 = 20 ? (AO[sup:2h9jmxfk]2[/sup:2h9jmxfk])(OB[sup:2h9jmxfk]2[/sup:2h9jmxfk])/4 = 400 ? (AO[sup:2h9jmxfk]2[/sup:2h9jmxfk])(OB[sup:2h9jmxfk]2[/sup:2h9jmxfk]) = 1600

? ((x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]/16)(16/(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16/(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]) = ((16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk])/16)((16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16)/(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]) = 1600

? ((16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk])(16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16) = 25600(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]

? 256(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk] + 256(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]6[/sup:2h9jmxfk] + 16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk] = 25600(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk]

? 16(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16 + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk] + (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] = (x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]4[/sup:2h9jmxfk] + 17(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16 = ((x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 16)((x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk] + 1) = 1600(x[sub:2h9jmxfk]A[/sub:2h9jmxfk])[sup:2h9jmxfk]2[/sup:2h9jmxfk]

I'm not sure whether I'm on the right track here, or where to go next.
 
You "seem" to be on right track...but your "typing presentation" gives me a headache!

Let the points be: A(u, u^2/4), B(v, v^2/4) and C(0,0)
So we have right triangle ABC with sides BC = a, AC = b and AB = c; so a^2 + b^2 = c^2.
STEP 1:
a^2 = v^2 + (v^2/4)^2 = (16v^2 + v^4) / 16 [1]

b^2 = u^2 + (u^2/4)^2 = (16u^2 + u^4) / 16 [2]

c^2 = (u + v)^2 + (u^2/4 - v^2/4)^4 = (16u^2 + 32uv + 16v^2 + u^4 - 2u^2v^2 + v^4) / 16 [3]

Since a^2 + b^2 = c^2, then [1] + [2] = [3], and this will simplify to uv = 16, so v = 16/u [4]

STEP 2:
We're given ab/2 = 20; so ab = 40; a^2b^2 = 1600 ; so [1] * [2] = 1600 :
[(16v^2 + v^4) / 16] * [(16u^2 + u^4) / 16] = 1600 ; simplifies to:
256u^2v^2 + 16u^4v^2 + 16u^2v^4 + u^4v^4 = 409600 [5]

Substitute [4] in [5], simplify to get this "cute quadratic !":
u^4 - 68u^2 + 256 = 0

Solve to get u = 2 or 8

So coordinates are (-2, 1) and (8, 16) OR (2, 1) and (-8, 16)

Are you impressed, Sir Khan? :roll:
 
Denis said:
You "seem" to be on right track...but your "typing presentation" gives me a headache!


Are you impressed, Sir Khan? :roll:
Click to expand...

I have been impressed since way back when .... that is why I call you for rescue...
 
\(\displaystyle Very \ good \ Denis, \ I'm \ impressed.\)

\(\displaystyle I \ also \ looked \ at \ this \ problem \ and \ on \ a \ scale \ of \ 1 \ to \ 10 \ decided \ that\)

\(\displaystyle it \ had \ an \ "Ick \ Factor" \ of \ 9.5, \ so \ I \ passed.\)

\(\displaystyle Actually, \ I \ would \ have \ worked \ on \ it, \ but \ was \ pressed \ for \ time.\)

\(\displaystyle However, your \ analysis \ of \ the \ problem \ was \ excellent, \ good \ show.\)
 
You must log in or register to reply here.
Top