Question

The number of points on the curve $$5x^{2} - 8xy + 5y^{2} = 4$$ whose distance from the origin is maximum or minimum is ________.

Solution

$$5x^{2}-8xy+5y^{2}=4$$$$(4x^{2}-4xy+y^{2})+(x^{2}-4xy+4y^{2})=4$$$$(2x-y)^{2}+(x-2y)^{2}=4$$$$\dfrac{2x-y)^{2}}{4}+\dfrac{(x-2y)^{2}}{4}=1$$From this equation we can see that the major axis of the ellipse is $$(2x-y=0)$$ and the minor axis $$(x-2y=0)$$two axis intersect at origin thus centre of ellipse is at originthus point doest to origin on the ellipse the point where the minor axis and ellipse intersectSimilarly the point farthest to origin on the ellipse on the ellipse the point where the major axis acid ellipse intersectLet the point lying on major axis be $$(a, b)$$$$(a, b)$$ also lies on line $$2x-y=0$$$$2a-b=0$$$$a=\dfrac{b}{2}$$ thus the point $$\left(\dfrac{b}{2}, b\right)$$also lies on ellipse$$(b-b)_{4}^{2}+\left(\dfrac{\dfrac{b}{2}-2b}{4}\right)^{2}=1$$$$\dfrac{9b^{2}}{4}=4\Rightarrow b=\pm \dfrac{4}{3}$$Hence the point $$(a,b)$$ is either$$\left(\dfrac{2}{3}, \dfrac{4}{3}\right)$$ or $$\left(\dfrac{-2}{3}, \dfrac{-4}{3}\right)$$ AnswerLet the point lying on minor axis be $$(c,d)$$ this point lies on line $$x=2y$$$$\therefore c=2d$$hence point becomes $$(2d, d)$$the point also satisfy equation of$$\dfrac{(4d-d)^{2}}{4}+\dfrac{(2d-2d)^{2}}{4}=1$$$$\dfrac{3d^{2}}{4}=1$$$$d=\pm \dfrac{2}{3}$$thus the point $$(c, d)$$ iseither $$\left(\dfrac{4}{3}, \dfrac{2}{3}\right)$$ or $$\left(\dfrac{-4}{3}, \dfrac{-2}{3}\right)$$Applied Mathematics

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