# ellipse problem

• Dec 19th 2008, 04:59 AM
varunnayudu
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 $\displaystyle x^2+2y^2=13.$
• Dec 19th 2008, 05:13 AM
Soroban
Hello, varunnayudu!

Quote:

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

Quote:

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.