2010 MATH destroyer #32 Test 1

[pandalove89]

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So the question is:


Two foci of an ellipse form the base of an isosceles triangle whose vertex coincides with a vertex of the ellipse as shown. If the equation of the ellipse is ((x^2)/(49)) + ((y^2)/(16)) = 1 what is the area of the triangle


I have no idea how to do this and even after reading the explanation it confused me even more.


I don't have a camera or screen shot of the figure so I'll try to explain it as best as I can.

Its an ellipse with x and y axis intersecting at the center. The isosceles triangle is sitting right above the x axis and its most upper vertex is hitting the top of the ellipse.

 

[indigenoustw]

First of all, find the distance from the origin to one of the foci (since the base of the triangle is formed between the foci).

If x^2/a^2 + y^2/b^2 = 1, let a = 7, b = 4, then foci = (c, 0) or (-c, 0)
One relationship to know: b^2 = a^2 - c^2 for the ellipse eqt givien above.

So solve for c, which equals to sqrt(49 - 16) = sqrt(33)

Now find the height of the triangle, which is (0, b), or 4.

Half of the isosceles triangle = 4*sqrt(33)/2 = 2*sqrt(33)

The whole isosceles triangle = 2*sqrt(33)*2 = 4 sqrt(33)



Hope it helps.

 

[LetsGo2DSchool]

Has anyone seen an ellipse problem on the DAT recently? I remember reading a score report thread the other day in which the person said there weren't any.

 

[toothhornet88]

Has anyone seen an ellipse problem on the DAT recently? I remember reading a score report thread the other day in which the person said there weren't any.

I would do an ellipse problem over a probability problem ANY day! lol