If the vertices of an ellipse are at (1, 5) and (1, –5) and (3, 0) is a point on the ellipse, what is the ellipse equation? (x^2)/(2^2) + (y – 1)^2/(5^2) = 1 (x – 1)^2/(2^2) + (y^2)/(5^2) = 1 (x^2)/(2^2) + (y^2)/(5^2) = 1 (x – 1)^2/(5^2) + (y^2)/(2^2) = 1
@freckles or @welshfella
http://www.wolframalpha.com/input/?i=%28x^2%29%2F%282^2%29+%2B+%28y^2%29%2F%285^2%29+%3D+1
aren't vertices the points
@freckles
If we make this traslation: \[\left\{ \begin{gathered} x = X + 1 \hfill \\ y = Y \hfill \\ \end{gathered} \right.\] then the equation of our ellipse, will be: \[\frac{{{X^2}}}{{{a^2}}} + \frac{{{Y^2}}}{{{b^2}}} = 1\] |dw:1429297793537:dw|
what is a?
2
yes!
sorry I have made an error, the right equation is: \[\frac{{{X^2}}}{{{b^2}}} + \frac{{{Y^2}}}{{{a^2}}} = 1\]
ok
so a is 5^2
or 5
what is b?
2?
starting from the old coordinates, x=3, y=0, we get the new coordinates X, Y using the trslation above, namely: X=2, Y=0 so substituting into the formula of the ellipse, we get: (4/b^2)+(0^2/25)=1 so b=2
traslation*
then the equation of our ellipse, in the XOY system is: (x^2/4)+(Y^2/25)=1
\[\frac{{{X^2}}}{4} + \frac{{{Y^2}}}{{25}} = 1\]
so its C
now, we have to rewrite that equation using the old coordinates, x, and y
so we have to replace X with x-1 and Y with y. So, what do you get?
(x – 1)^2/(2^2) + (y^2)/(5^2) = 1
please you have to keep in mind the traslation above. So yes! that's right!
so its B
yes!
thanks
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