Answer:
4733
Step-by-step explanation:
Please refer to the attached diagram.
Point A can be assigned x-coordinate "p". Then its y-coordinate is 6p^2. The slope at that point is y'(p) = 12p.
Point B can be assigned x-coordinate "r". Then its y-coordinate is 6r^2. The slope at that point is y'(r) = 12r.
We want the slopes at those points to have a product of -1 (so the tangents are perpendicular). This means ...
(12p)(12r) = -1
r = -1/(144p)
The slope of line AB in the diagram is the ratio of the differences of y- and x-coordinates:
slope AB = (ry -py)/(rx -px) = (6r^2 -6p^2)/(r -p) = 6(r+p) . . . . simplified
The slope of AB is also the tangent of the sum of these angles: the angle AC makes with the x-axis and angle CAB. The tangent of a sum of angles is given by ...
tan(α+β) = (tan(α) +tan(β))/1 -tan(α)·tan(β))
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Of course the slope of a line is equal to the tangent of the angle it makes with the x-axis. The tangent of angle CAB is 2 (because the aspect ratio of the rectangle is 2). This means we can write ...
slope AB = ((slope AC) +2)/(1 -(slope AC)(2))
[tex]6(p+r)=\dfrac{12p+2}{1-(12p)(2)}\\\\3(p+r)(1-24p)=6p+1\qquad\text{multiply by $1-24p$}\\\\3\left(p-\dfrac{1}{144p}\right)(1-24p)=6p+1\qquad\text{use the value for r}\\\\3(144p^2-1)(1-24p)=144p(6p+1)\qquad\text{multiply by 144p}\\\\ 3456 p^3+ 144 p^2+ 24 p+1 =0\qquad\text{put in standard form}\\\\144p^2(24p+1)+(24p+1)=0\qquad\text{factor by pairs}\\\\(144p^2+1)(24p+1)=0\qquad\text{finish factoring}\\\\p=-\dfrac{1}{24}\qquad\text{only real solution}\\\\r=\dfrac{-1}{144p}=\dfrac{1}{6}[/tex]
So, now we can figure the coordinates of points A and B, and the distance between them. That distance is given by the Pythagorean theorem as ...
d^2 = (6r^2 -6p^2)^2 +(r -p)^2
d^2 = (6(1/6)^2 -6(-1/24)^2)^2 +(1/6 +1/24)^2 = 25/1024 +25/576 = 625/9216
Because of the aspect ratio of the rectangle, the area is 2/5 of this value, so we have ...
Rectangle Area = (2/5)(625/9216) = 125/4608 = a/b
Then a+b = 125 +4608 = 4733.
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Comment on the solution
The point of intersection of the tangent lines is a fairly messy expression, and that propagates through any distance formulas used to find rectangle side lengths. This seemed much cleaner, though maybe not so obvious at first.
