• January 2nd 2008, 10:12 AM
rlarach
:confused:The graphs of x/4 + y/3 = 1 and (x^2)/16 + (y^2)/9 = 1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3.

Thank you very much for your help!! I appreciate it! :):)
• January 2nd 2008, 11:24 PM
mr fantastic
Quote:

Originally Posted by rlarach
:confused:The graphs of x/4 + y/3 = 1 and (x^2)/16 + (y^2)/9 = 1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3.

Thank you very much for your help!! I appreciate it! :):)

Since no-one else has replied I'll suggest an approach:

1. Go to parametric equation for the ellipse: x = 4 cos t, y = 3 sin t. Then the vertices of the triangle are A(0, 3), B(4, 0) and C(4 cos t, 3 sin t).

2. The area of a triangle whose vertices have coordinates (x1, y1), (x2, y2) and (x3, y3) is A = 1/2 |(x1 - x3)(y1- y3) - (y1 - y2)(x1 - x3)| (you should prove this - only takes a couple of lines).

3. Sub in the coordinates of the vertices of the triangle and solve A = 3. You get a trig equation that you may need help solving.

There's probably a much more elegant approach - maybe it's no coincidence that if you consider the point D(0, -3) then triangle ABD has an area of 12 = 4 x 3 .....
• January 3rd 2008, 12:28 PM
topsquark
Quote:

Originally Posted by mr fantastic
There's probably a much more elegant approach

Actually, since I tend to "bull" my way through a problem, I thought your solution was quite elegant.

-Dan
• January 4th 2008, 11:37 AM
mr fantastic
Quote:

Originally Posted by topsquark
Actually, since I tend to "bull" my way through a problem, I thought your solution was quite elegant.

-Dan

Ta. Actually I tend to bull$&# my way through a problem (Rofl) (You know what they say .... If you can't blind them with you brilliance, baffle them with you bull$&# (Rofl)
• January 4th 2008, 11:40 AM
colby2152
Quote:

Originally Posted by mr fantastic
Ta. Actually I tend to bull$&# my way through a problem (Rofl) (You know what they say .... If you can't blind them with you brilliance, baffle them with you bull$&# (Rofl)

I had to thank you for saying one of my favorite lines!

"If you can't blind them with you brilliance, baffle them with you bull$&#" -Anonymous • January 4th 2008, 11:56 AM topsquark Quote: Originally Posted by mr fantastic (You know what they say .... If you can't blind them with you brilliance, baffle them with you bull$&# (Rofl)

Hmmmm... Sounds like you attended my Thesis presentation for my MS.

-Dan
• January 4th 2008, 08:17 PM
Isomorphism
Quote:

Originally Posted by rlarach
:confused:The graphs of x/4 + y/3 = 1 and (x^2)/16 + (y^2)/9 = 1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3.
Thank you very much for your help!! I appreciate it! :):)

There is another approach(a little easier in finding the area of the triangle)

First off, Mr.F's (1) says (4,0) and (0,3) are the points...

Now I assume you know the perpendicular distance of a point (m,n) from the line Ax+By+C = 0 is $\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}$----(*)

From this it is a tad easier because $AB = \sqrt{(3-0)^2+(0-4)^2} = 5$
From
and height(from (*)) is $\frac{|\frac{m}{4}+\frac{n}{3}-1|}{\sqrt{(\frac14)^2+(\frac13)^2}}$
and (m,n) lies on the ellipse ,so $(m,n) = (4 \cos t, 3 \sin t)$
Now, $Area = \frac12 \cdot base \cdot height$
Base = AB = 5
Height = $\frac{|\frac{4 \sin t}{4}+\frac{3\cos t}{3}-1|}{\sqrt{(\frac14)^2+(\frac13)^2}}= \frac{12}{5} \cdot |\sin t + \cos t - 1|$
You want A to be 3, which means
$|\sin t + \cos t - 1| = \frac12$
From now on, it is just basic trigonometry
This is not less messier than Mr.Fs way, but a different approach :)