I'm preparing for the ACTM Sate contest, and I stumbled across this problem. I've never learned how to decipher equations like this, so could someone explain what these equations mean and how to solve this problem?
The area of an ellipse is calculated using the formula $A = \pi*a*b$, where a is half the length of the major axis of the ellipse and b is half the length of the minor axis of the ellipse. What is the area enclosed by $9x^2 + 25y^2 + 72x - 250y + 544 = 0$?
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$\begingroup$$$9(x+4)^2+25(y-5)^2=144+625-544=15^2$$
$$\implies\dfrac{(x+4)^2}{(15/3)^2}+\dfrac{(y-5)^2}{(15/5)^2}=1$$
Can you recognize $a,b$ here?
The equation for the area of an ellipse is $A=\pi ab$
You can see that $a$ is the horizontal radius, and $b$ is the vertical radius.
$\endgroup$ 6 $\begingroup$There is no $xy$ term but there are linear terms, so this ellipse is unrotated, but is translated from standard position. We can find the center by differentiating: $$\nabla(9x^2+25y^2+72x-250y+544)=\langle18x+72,50y-250\rangle=0,$$ therefore the center is at $(-4,5)$. Translating will eliminate the linear terms from the equation while leaving the coefficients of the quadratic terms unchanged, and the new constant term can be found by plugging the center coordinates into the left-hand side of the ellipse’s equation: $$9(-4)^2+25(5)^2+72(-4)-250(5)+544=-255=-(15)^2.$$ The equation of the ellipse can thus be rewritten as $$3^2(x+4)^2+5^2(y-5)2=15^2,$$ and we can find $a$ and $b$ by dividing by the right-hand side of this equation: $a=15/3=5$ and $b=15/5=3$.
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