The ellipse with x-intercepts (5, 0) and (-5, 0); y-intercepts (0, 3) and (0, -3). https://hanwingspanps.owschools.com/media/g_alg02_ccss_2015/8/l7elpseq.gif i need this put into this equation
cool stuff we have the center is (0,0) then a is just the length between (0,0) and (5,0) (or the length between (0,0) and (-5,0)) b is just the length between (0,0) and (0,3) ( or the length between (0,0) and (0,-3))
what is x and y then
x and y are variables
without the variables we wouldn't be able to make an ellipse (x,y) is the point that varies around the ellipse
the only values you need to find are a and b
but x,y are points already given ?
because i need to completely fill in the equation
x-intercepts is where the ellipse pass through the x-axis this is where the y-coordinate will be 0 y-intercepts is wehre the ellipse pass through the y-axis this is where the x-coordinate will be 0 y and x aren't always 0... I'm trying to say there x and y are variables where as the x-intercepts and y-intercepts are ordered pairs where both coordinates are constant value
the only values you need to find are a and b
x and y stay as they are because they are variables
so i should plot the lnes than =0
no
we have the center is (0,0) then a is just the length between (0,0) and (5,0) (or the length between (0,0) and (-5,0)) b is just the length between (0,0) and (0,3) ( or the length between (0,0) and (0,-3)) this is all you need to know
find a and b using this a is the length between the center and the x-intercept b is the length between the center and the y-intercept in order words
how far is 5 from 0? how far is 3 from 0?
a=10 b=6 or a=5 b=3
where do you get a=10 or b=6? 5 is only 5 units from 0 and 3 is only 3 units from 0
so a=5 and b=3
im going 5 from -5
yeah we don't want to do that... I can show you a more convincing way I think earlier you so badly wanted to plug in our x-intercept and y-intercepts we can do that... and we can use that to find a and b \[(5,0) \text{ inset into } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \frac{5^2}{a^2}+\frac{0^2}{b^2}=1 \\ \frac{5^2}{a^2}=1 \\ 5^2=a^2 \\ a^2=25 \] you can use (-5,0) to find a^2 to be 25 also
\[(0,3) \text{ insert into } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \frac{0^2}{a^2}+\frac{3^2}{b^2}=1 \\ \frac{3^2}{b^2}=1 \\ 3^2=b^2 \\ b^2=9\]
the ellipse is therefore \[\frac{x^2}{25}+\frac{y^2}{9}=1 \]
a=5 which is the distance from the x-intercept to the center b=3 which is the distance from the y-intercept to the center
by showing me the awnser i understand it ten fold
you don't have to do all of that math is what I'm saying but you can
okay so you only doing from one set point to center not both
for example here is another problem: The ellipse with x-intercepts (2,0) and (-2,0); y-intercepts (0,9) and (0,-9) the distance from x-intercept to center is 2 this is your a the distance from y-intercept to center is 9 this is your b
but you could do some algebra like I did above to find this
@mathmale close your eyes ive betrayed you
\[\frac{x^2}{2^2}+\frac{y^2}{9^2}=1 \] would be the answer to my example
do you understand how I got a=2 and b=3?
for my example that is
okay i get it........... thanks once again
np
It would help so much if you, DD, would sketch this parabola. Sketch the four given points, each of which represents a vertex. Once you've completed this sketch, we can further discuss what a and b mean and where they come from. Please try to understand the example given you by freckles; it's accurate. freckles Best Response Medals 1 for example here is another problem: The ellipse with x-intercepts (2,0) and (-2,0); y-intercepts (0,9) and (0,-9) the distance from x-intercept to center is 2 this is your a the distance from y-intercept to center is 9 this is your b
its been a long day.....i really wish i could thank you in person ......everything and everyone on here is so generous and helpful
The ellipse with x-intercepts at (4, 0) and (-4, 0), y-intercepts at (0, 9) and (0, -9), and center at (0, 0). a^2=20 b^2=81
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