Please help with one problem about writing ellipse in standard form?

[JakeyTheIdiot]

Write the equation for an ellipse with vertices (0,-3) and (0,3), minor axis of length 10.

I know that the standard form of an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2=1

Please help me ! Please!

Thank you so much for your time, I appreciate it.

 

[MarkFL]

It appears that the minor axis is longer than the major axis. Are you certain you have copied the problem correctly?

 

[JakeyTheIdiot]

Yes, I am certain :p I think the minor axis is the vertical length of the circle

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It appears that the minor axis is longer than the major axis. Are you certain you have copied the problem correctly?

Yes, I am positive! :P

 

[MarkFL]

Write the equation for an ellipse with vertices (0,-3) and (0,3), minor axis of length 10.

I know that the standard form of an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2=1

Please help me ! Please!

Thank you so much for your time, I appreciate it.

The posted problem is impossible, so let's look at the problem:

Write the equation for an ellipse with co-vertices $(0,-3)$ and $(0,3)$, major axis of length 10.

Can you identify where the center of the ellipse must be?

 

[JakeyTheIdiot]

The posted problem is impossible, so let's look at the problem:

Write the equation for an ellipse with co-vertices $(0,-3)$ and $(0,3)$, major axis of length 10.

Can you identify where the center of the ellipse must be?

(h,k) = (0,0) so the center is (0,0).

 

[MarkFL]

(h,k) = (0,0) so the center is (0,0).

That is the correct center, but the reason is that the center is the mid-point of the co-vertices:

\(\displaystyle (h,k)=\left(\frac{0+0}{2},\frac{-3+3}{2}\right)=(0,0)\)

So, given that the major axis is perpendicular to the minor axis at the center, can you give the coordinates of the vertices?

 

[JakeyTheIdiot]

That is the correct center, but the reason is that the center is the mid-point of the co-vertices:

\(\displaystyle (h,k)=\left(\frac{0+0}{2},\frac{-3+3}{2}\right)=(0,0)\)

So, given that the major axis is perpendicular to the minor axis at the center, can you give the coordinates of the vertices?

but the thing is the vertices are already given (0,-3) and (0,3) that's why I am so confused , I think it might be the vertical kind of ellips with an equation like x^2/b^2 +y^2/a^2=1 but I have no clue what to sub in.

 

[MarkFL]

but the thing is the vertices are already given (0,-3) and (0,3) that's why I am so confused , I think it might be the vertical kind of ellips with an equation like x^2/b^2 +y^2/a^2=1 but I have no clue what to sub in.

We were given the co-vertices...the end-points of the shorter axis (it has length 6). We want to find the end-points now of the major axis (the longer axis, having length 10), which are called the vertices. :D

 

[JakeyTheIdiot]

We were given the co-vertices...the end-points of the shorter axis (it has length 6). We want to find the end-points now of the major axis (the longer axis, having length 10), which are called the vertices. :D

Ohhh I see a^2 = 6 so the vertices are (0,-6) and (0,6)

 

[MarkFL]

Ohhh I see a^2 = 6 so the vertices are (0,-6) and (0,6)

No, the slope of the minor axis is:

\(\displaystyle m=\frac{-3-3}{0-0}=\text{undefined}\)

It is a vertical line, so the major axis must be horizontal, and since it must pass through the center, which we determined is the origin, the major axis must therefore lie along the $x$-axis.

So, we may say the vertices are at:

\(\displaystyle (\pm v,0)\) where $0<v$

Now, we require the distance between these vertices to be 10 units, so we may write:

\(\displaystyle 10=v-(-v)=2v\implies v=5\)

Hence, the vertices are at:

\(\displaystyle (\pm5,0)\)

So, what must $a$ and $b$ be?

 

[JakeyTheIdiot]

No, the slope of the minor axis is:

\(\displaystyle m=\frac{-3-3}{0-0}=\text{undefined}\)

It is a vertical line, so the major axis must be horizontal, and since it must pass through the center, which we determined is the origin, the major axis must therefore lie along the $x$-axis.

So, we may say the vertices are at:

\(\displaystyle (\pm v,0)\) where $0<v$

Now, we require the distance between these vertices to be 10 units, so we may write:

\(\displaystyle 10=v-(-v)=2v\implies v=5\)

Hence, the vertices are at:

\(\displaystyle (\pm5,0)\)

So, what must $a$ and $b$ be?

wait so does that mean a^2 is the minor axis squared so itd be 36? I`m not sure but I think so

 

[MarkFL]

We know that since the center of the ellipse is at the origin, then the equation of the ellipse will have the form:

\(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Now, when $x=0$, we then have:

\(\displaystyle \frac{y^2}{b^2}=1\)

\(\displaystyle y^2=b^2\implies y=\pm b\)

What two points on the ellipse were we given with $x=0$?