ellipse (1 Viewer)

Masaken

Member
are ellipses anywhere in the extension 1 course (+ more precisely, year 11 in the new syllabus)? wondering because a maths qn i'm doing is asking me to sketch an ellipse after finding its equation and i have no idea how to do it

yanujw

Active Member
Technically it could be done by a dilation of a circle, although I'd expect that to be within the 2u course. Perhaps upload the question for us to look at?

Masaken

Member
Technically it could be done by a dilation of a circle, although I'd expect that to be within the 2u course. Perhaps upload the question for us to look at?
A curve is defined parametrically by the equations x = cos theta + 1, y = 2sintheta - 3.
(a) Find the Cartesian equation of the curve.
(b) Describe the curve, and then sketch it.

I ended up getting (x-1)^2 + [(y+3)^2/4] = 1 for the Cartesian equation, and while I know it's an ellipse I've no idea where to start with sketching it

5uckerberg

Well-Known Member
Okay do you mean $\bg_white \frac{\left(x-1\right)^{2}}{1}+\frac{\left(y+3\right)^{2}}{4}=1$.

Okay first and foremost for this question we are going to start with the basics okay the equation for the circle is $\bg_white x^{2}+y^{2}=r^{2}$.
Well, we want to turn either $\bg_white \left(x-1\right)^{2}$ or $\bg_white \left(y+3\right)^{2}$ into zero first then we can start talking.

Knowing that let $\bg_white x=1$ then we have $\bg_white \frac{\left(y+3\right)^{2}}{4}=1$. Next step we have $\bg_white \left(y+3\right)^{2}=4$ then $\bg_white y+3=\pm{2}$ therefore, $\bg_white y=-5, -1$.

Do the same for $\bg_white y=-3$ then you will see that $\bg_white \left(x-1\right)^{2}=1$. Then of course $\bg_white x-1=\pm{1}$ there $\bg_white x=0, 2$

At this point draw up the points $\bg_white \left(0, 3\right) \left(2, 3\right) \left(1, -5\right) \left(1, -1\right)$ and then trace out the ellipse.

5uckerberg

Well-Known Member
Please note learn to type your equation in latex thank you it was a bit hard to read at first because I thought it was $\bg_white \left(x-1\right)^{2}+\sqrt{y+3}$ that you were found.

Your problem if typed in Latex would have to be done like this

1. Press f(x) which will bring up LaTeX
2. Type in \left(x-1\right)^{2}
3. Type in \frac{\left(y+3\right)^{2}}{4}
4. Type in = 1
5. Completed.

Masaken

Member
Please note learn to type your equation in latex thank you it was a bit hard to read at first because I thought it was $\bg_white \left(x-1\right)^{2}+\sqrt{y+3}$ that you were found.

Your problem if typed in Latex would have to be done like this

1. Press f(x) which will bring up LaTeX
2. Type in \left(x-1\right)^{2}
3. Type in \frac{\left(y+3\right)^{2}}{4}
4. Type in = 1
5. Completed.
Thanks a lot, will take note of that in the future