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# conic sections watch

1. I don't really have an idea of what to do, so an example or an explanation for one question will be good. Thanks.

2. The following equations describe conic sections in the (x, y) coordinate plane. In each
case sketch the curve, showing its main features clearly, and calculate its eccentricity.
(a) (y + 1)^2 + 7(x − 4) = 0
(b) 16(x + 1)^2 − 8(y − 3)^2 − 16 = 0
(c) (y + 1) − 7(x − 1) = 0
(d) 3(x + 2)^2 + 3(y − 1)^2 − 27 = 0
(e) (x − 1)^2 + 2(y − 2)^2 + 4 = 0

Also, if anyone can sketch graphs on here, can you confirm what r = a cos 2(theta) looks like please. That would be really helpful thanks.
2. (Original post by barry_4_england)
2. The following equations describe conic sections in the (x, y) coordinate plane. In each
case sketch the curve, showing its main features clearly, and calculate its eccentricity.
(a) (y + 1)^2 + 7(x ? 4) = 0
(b) 16(x + 1)^2 ? 8(y ? 3)^2 ? 16 = 0
(.
Here are two graphs.
Attached Images

3. Cheers, but how do you work out how to draw them from the equations? And calculate the eccentricity?
4. (Original post by barry_4_england)
I don't really have an idea of what to do, so an example or an explanation for one question will be good. Thanks.

2. The following equations describe conic sections in the (x, y) coordinate plane. In each
case sketch the curve, showing its main features clearly, and calculate its eccentricity.
(a) (y + 1)^2 + 7(x ? 4) = 0
(b) 16(x + 1)^2 ? 8(y ? 3)^2 ? 16 = 0
(c) (y + 1) ? 7(x ? 1) = 0
(d) 3(x + 2)^2 + 3(y ? 1)^2 ? 27 = 0
(e) (x ? 1)^2 + 2(y ? 2)^2 + 4 = 0

Also, if anyone can sketch graphs on here, can you confirm what r = a cos 2(theta) looks like please. That would be really helpful thanks.
Area you sure all your equations are typed out correctly ?

I know that a circle has the general equation (x - a)^2 + (y - b)^2 = r^2 so I would expect conic sections to have similar equations, with a coefficient infront of at least one bracket.

In your equation (c) you don't have any powers - so this would be a straight line.
5. (Original post by barry_4_england)
.
Here I have displayed two graphs side by side.
Attached Images

6. Well yer d) will be a circle (-2,1) radius 3, but a circle is a conic sections, as it is an ellipse with 2 focus points equal to eachother. I'm just confused on how to work out the eccentricity we haven't really done much on these yet.

https://learn.lboro.ac.uk/ma/06maa14...mymodules.html

7. (a) (y + 1)^2 + 7(x − 4) = 0 Parabola
(b) 16(x + 1)^2 − 8(y − 3)^2 − 16 = 0 Hyperbola
(c) (y + 1) − 7(x − 1) = 0 Line
(d) 3(x + 2)^2 + 3(y − 1)^2 − 27 = 0 Circle
(e) (x − 1)^2 + 2(y − 2)^2 + 4 = 0 Nothing. (two positive numbers cannot add to give -4) I assume that should be -4. In which case it's an ellipse.

To sketch a) think about a shift in the coordinate axes to get it looking like something you can draw. Let Y=y+1 and X=x-4. So the curve is Y^2 = -7X. Which is in a standard form (which you should be able to draw. Don't forget that it's reflected in the Y axis as there's a negative number infront of X). Then shift the curve back. So putting Y=y+1 back will move the curve down 1 unit and putting in X=x-4 will move it right 4 units. It should then be easy enough to label where it crosses the axes. Do the others you're having trouble drawing in a similar way.

Now note that a shift of the axes doesn't change the eccentricity. So using substitutions like above, get the equations in b and e into the following forms:
Hyperbola:
(X/a)^2 - (Y/b)^2 = 1
Ellipse:
(X/a)^2 + (Y/b)^2 = 1

8. Eccentricity for hyperbola http://en.wikipedia.org/wiki/Hyperbola
Attached Images

9. Thanks for you help guys, I'll give it a go.

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