ellipse problem

**varunnayudu** · December 19th 2008, 04:59 AM

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

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

**Soroban** · December 19th 2008, 05:13 AM

Hello, varunnayudu!

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

Substitute $(\alpha,\beta)$ into the equation.
Compare the left side to the right side.

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

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

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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