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Thread: ellipse problem

  1. #1
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    ellipse problem

    How will you find out if the point $\displaystyle (\alpha\beta)$ is outside, on or inside the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

    Hence, show that the triangle whose vertices are $\displaystyle (1,2),(3,-1)$ and $\displaystyle (-2,1)$ lies wholly inside the ellipse $\displaystyle x^2+2y^2=13.$
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  2. #2
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    Hello, varunnayudu!

    How will you find out if the point $\displaystyle (\alpha,\beta)$ is outside, on or inside the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}\:=\:1$
    Substitute $\displaystyle (\alpha,\beta)$ into the equation.
    Compare the left side to the right side.

    If the left is: .$\displaystyle \begin{Bmatrix}\;< & \text{inside} \\ \;= & \text{on} \\ \;> & \text{outside} \end{Bmatrix}$




    Hence, show that the triangle whose vertices are $\displaystyle (1,2),(3,-1)$ and $\displaystyle (-2,1)$
    lies wholly inside the ellipse $\displaystyle x^2+2y^2\:=\:13$
    Apply the above test to the three vertices.

    Since an ellipse is convex, if the vertices are inside the ellipse,
    . . the sides are also inside the ellipse.

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