Maximizing Area of an Ellipse Passing Through a Fixed Point

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The discussion focuses on finding the equation of an ellipse that is symmetric about the coordinate axes and passes through a fixed point (h, k) while minimizing its area. The equation of the ellipse is given by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, and the area is expressed as $\pi ab$. The condition for the point (h, k) to lie on the ellipse is $\dfrac{h^2}{a^2} + \dfrac{k^2}{b^2} = 1$. The method of Lagrange multipliers is suggested for solving the minimization problem, although a parametric approach is also discussed due to the audience's limited exposure to multivariable calculus.

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347) An ellipse slmetric with respect to the coordinate axes handle: through the fixed point (h, k). Find The equation of the ellipse's?
area maximum
answer k2/SUP] h 2 + h 2 y 2 = 2 h 2 k2 The equations here must be
y = m(x-h)+k

Parametrizing
x= a cos(t)
y = b(sen(t)

D= (a cost-h)2+((macosth-h)2+ k -k)2

am I right?
Becuse I could not get the right answer
 
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leprofece said:
347) An ellipse slmetric with respect to the coordinate axes handle: through the fixed point (h, k). Find The equation of the ellipse's?
area maximum
answer k2/SUP] h 2 + h 2 y 2 = 2 h 2 k2 The equations here must be
y = m(x-h)+k

Parametrizing
x= a cos(t)
y = b(sen(t)

D= (a cost-h)2+((macosth-h)2+ k -k)2

am I right?
Becuse I could not get the right answer

If I understand it correctly, the problem is to find an ellipse symmetric about the coordinate axes and passing through the fixed point $(h,k)$, such that the ellipse has the minimum possible area. (The question says maximum area, but in fact the only critical point is a minimum, and the maximum area is unbounded.)

The equation of the ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, and we have to choose $a$ and $b$ so as to minimise the area of the ellipse, which is $\pi ab.$ The condition for the point $(h,k)$ to lie on the ellipse is $\dfrac{h^2}{a^2} + \dfrac{k^2}{b^2} = 1$, or $a^2k^2 + b^2h^2 = a^2b^2$. So we want to minimise $ab$ subject to the constraint $a^2k^2 + b^2h^2 = a^2b^2$. The usual way to solve problems of that sort is to use the method of Lagranga multipliers.
 
Opalg said:
If I understand it correctly, the problem is to find an ellipse symmetric about the coordinate axes and passing through the fixed point $(h,k)$, such that the ellipse has the minimum possible area. (The question says maximum area, but in fact the only critical point is a minimum, and the maximum area is unbounded.)

The equation of the ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, and we have to choose $a$ and $b$ so as to minimise the area of the ellipse, which is $\pi ab.$ The condition for the point $(h,k)$ to lie on the ellipse is $\dfrac{h^2}{a^2} + \dfrac{k^2}{b^2} = 1$, or $a^2k^2 + b^2h^2 = a^2b^2$. So we want to minimise $ab$ subject to the constraint $a^2k^2 + b^2h^2 = a^2b^2$. The usual way to solve problems of that sort is to use the method of Lagranga multipliers.

thanks sorry but the students here haven't studied 2wo variables yet so it should be answerred or with one variable or with parametrics So I wrote parametrics in what I did
can you or anyone help me ??
 
Perhaps you could solve the constraint for either of the two variables and then use substitution to obtain an objective function in one variable...(Thinking)
 

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