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A chord of negative slope from the point P(${\sqrt{264}, 0)}$ is drawn to the ellipse ${x^2} +4 {y^2}=16$ . This chord intersects the ellipse at A and B.(O is the origin). Find the maximum area of the $\triangle$ AOB and the slope of the line AB when the area is maximum.

My trial ------ Let the chord intersect the ellipse at two points (4cos(${\alpha}$),2sin(${\alpha}$)) and (4cos(${\beta}$),2sin(${\beta}$)) This would lead to the chord equation ----> $\frac{x}{4}$$cos(\frac{\alpha+\beta}{2})$ + $\frac{y}{2}$sin($\frac{\alpha+\beta}{2})$= cos($\frac{\alpha-\beta}{2})$. This gives $\frac{\sqrt{264}}{4}$$cos(\frac{\alpha+\beta}{2})$= cos($\frac{\alpha-\beta}{2})$. And I've found length of OA and OB.... and tried differentiating to get maximum area. But it is ending up in vain and more over too lengthy. The answer given is maximum area =4, and slope of the line AB is ($\frac{-1}{8\sqrt{2}})$. Any help/hint will be appreciated. Thanks in advance

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Let $OA$ be a semi-diameter of the ellipse and the tangent at $B$ be parallel to it. It is evident (see figure below) that the area of any other triangle with the same base $OA$ and its third vertex on the ellipse cannot be greater tan the area of triangle $OAB$.

Two semi-diameters like $OA$ and $OB$ are called conjugate semi-diameters and by the second theorem of Apollonius the area of a triangle formed by any two conjugate semi-diameters is always ${1\over2}ab$, where $a$ and $b$ are the semi-axes of the ellipse.

Hence no calculus is needed to answer the given question: maximum area is ${1\over2}ab=4$ and the chord must be drawn so that points $A$ and $B$ are the endpoints of two conjugate semi-diameters. To this end it is useful to know that if $A=(4\cos\alpha,2\sin\alpha)$ then $OB$ is conjugate to $OA$ if $$B=\big(4\cos(\pi/2+\alpha),2\sin(\pi/2+\alpha)\big)=(-4\sin\alpha,2\cos\alpha).$$ The value of $\alpha$ can then be found by imposing line $AB$ to pass through point $P$.

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