# Ellipse problem

• Jan 7th 2010, 07:19 PM
thanapa
Ellipse problem
How will u find out that 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 verities are (1,2), (3,-11) and (-2,1) lies wholly inside the ellipse $\displaystyle x^2 + 2y^2 = 13.$
• Jan 7th 2010, 07:20 PM
Prove It
Quote:

Originally Posted by thanapa
How will u find out that 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 verities are (1,2), (3,-11) and (-2,1) lies wholly inside the ellipse $\displaystyle x^2 + 2y^2 = 13.$

This entire problem can be solved by graphing.
• Jan 7th 2010, 07:21 PM
thanapa
Quote:

Originally Posted by Prove It
This entire problem can be solved by graphing.

but i need some steps ..if u could provide any
• Jan 7th 2010, 08:11 PM
Roam
Quote:

Originally Posted by thanapa
but i need some steps ..if u could provide any

Look at the standard form equation of the ellipse: $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

if the constant a>b it represents the length of the major axis, and if a<b it represents the minor axis. The same thing applies to "b".

You can now sketch the ellipse. But you can also check this without drawing a graph, if you have studied the quadratic forms.
• Jan 7th 2010, 08:16 PM
Shanks
substitute$\displaystyle (x,y)$ with $\displaystyle (\alpha,\beta)$, If the value is less than 1, then inside; equal to 1, lie on the curve; is greater than 1, outside.