Sketching Hyperbola over complex plane (1 Viewer)

[clintmyster]

My teacher told us to sketch: |z + 10| - |z - 10| = 12

Now we did it without letting z = x + iy and sqrrooting and so forth. We tried to do it geometrically. If you recall sketching an ellipse like |z - 5| + |z + 5| = 26, the foci lie at 5,0 and -5,0 and the length of the major axis PS + PS' = 2a = 26 hence a = 13, e = 5/13 and you can work out b to be 12. Now for a hyperbola he is like you use PS' - PS = 2a however this implies that the value of PS' is greater then PS and therefore PS' lies further away from the hyperbola, therefore you sketch the RIGHT side of a hyperbola. He also said if the question was |z + 5| + |z - 5| = 26 then PS - PS' = 2a and that PS > PS' and therefore PS lies further away from the hyperbola therefore you sketch the LEFT side of a hyperbola.

Now i thought if this were true then wouldnt your eqn of the hyperbola alter somehow like for the top part of a circle, you have y=sqrroot (a² - x²). But he seemed to have the same eqn for the hyperbola so I got confused why is this the case. I tried asking him to explain this whole thing again to me but he said the same thing so I decided to come here. He said this stuff is in no textbook however sketching the ellipse in this way is in the complex numbers section of patel (but im really more interested in the hyperbola).

Can anyone please clarify this with me? If you need diagrams i'l scan them. Thanks

 

[Drongoski]

My teacher told us to sketch: |z + 10| - |z - 10| = 12

Now we did it without letting z = x + iy and sqrrooting and so forth. We tried to do it geometrically. If you recall sketching an ellipse like |z - 5| + |z + 5| = 26, the foci lie at 5,0 and -5,0 and the length of the major axis PS + PS' = 2a = 26 hence a = 13, e = 5/13 and you can work out b to be 12. Now for a hyperbola he is like you use PS' - PS = 2a however this implies that the value of PS' is greater then PS and therefore PS' lies further away from the hyperbola, therefore you sketch the RIGHT side of a hyperbola. He also said if the question was |z + 5| + |z - 5| = 26 then PS - PS' = 2a and that PS > PS' and therefore PS lies further away from the hyperbola therefore you sketch the LEFT side of a hyperbola.

Now i thought if this were true then wouldnt your eqn of the hyperbola alter somehow like for the top part of a circle, you have y=sqrroot (a² - x²). But he seemed to have the same eqn for the hyperbola so I got confused why is this the case. I tried asking him to explain this whole thing again to me but he said the same thing so I decided to come here. He said this stuff is in no textbook however sketching the ellipse in this way is in the complex numbers section of patel (but im really more interested in the hyperbola).

Can anyone please clarify this with me? If you need diagrams i'l scan them. Thanks[/qu


One way of characterising a hyperbola is that the difference of the distances of any point on it from the 2 foci (in this case (-10, 0) & (10,0) ) is +/- a constant (12 in this case); if 1 distance is 5 and the other 17, then 5 - 17 would be -12 rather than 12.

So in this case 12 = 2a and therefore since 10 = ae and a = 6, e = 10/6 = 5/3 and therefore, from b^2 = a^2(e^2 - 1) we get b = 8
Therefore the locus is the hyperbola x^2/6^2 - y^2/8^2 = 1

Hope that's right and hopefully provides some clarification.

[Observation: If a teacher is good and knows his stuff, and you ask him to explain again, he would give you an alternative explanation or two to help you understand. If a teacher is not in mastery of a topic, it is easy enough for him to provide an example or solution taken from another source but which he cannot solve himself I've sometimes heard students complaining that when asked to explain again when they still cannot understand, the teacher repeats the explanation almost word for word; this usually means the teacher did not do the solution himself and would be frustrated if you keep badgering him to explain.]

 

[clintmyster]

My teacher told us to sketch: |z + 10| - |z - 10| = 12

Now we did it without letting z = x + iy and sqrrooting and so forth. We tried to do it geometrically. If you recall sketching an ellipse like |z - 5| + |z + 5| = 26, the foci lie at 5,0 and -5,0 and the length of the major axis PS + PS' = 2a = 26 hence a = 13, e = 5/13 and you can work out b to be 12. Now for a hyperbola he is like you use PS' - PS = 2a however this implies that the value of PS' is greater then PS and therefore PS' lies further away from the hyperbola, therefore you sketch the RIGHT side of a hyperbola. He also said if the question was |z + 5| + |z - 5| = 26 then PS - PS' = 2a and that PS > PS' and therefore PS lies further away from the hyperbola therefore you sketch the LEFT side of a hyperbola.

Now i thought if this were true then wouldnt your eqn of the hyperbola alter somehow like for the top part of a circle, you have y=sqrroot (a² - x²). But he seemed to have the same eqn for the hyperbola so I got confused why is this the case. I tried asking him to explain this whole thing again to me but he said the same thing so I decided to come here. He said this stuff is in no textbook however sketching the ellipse in this way is in the complex numbers section of patel (but im really more interested in the hyperbola).

Can anyone please clarify this with me? If you need diagrams i'l scan them. Thanks[/qu


One way of characterising a hyperbola is that the difference of the distances of any point on it from the 2 foci (in this case (-10, 0) & (10,0) ) is +/- a constant (12 in this case); if 1 distance is 5 and the other 17, then 5 - 17 would be -12 rather than 12.

So in this case 12 = 2a and therefore since 10 = ae and a = 6, e = 10/6 = 5/3 and therefore, from b^2 = a^2(e^2 - 1) we get b = 8
Therefore the locus is the hyperbola x^2/6^2 - y^2/8^2 = 1

Hope that's right and hopefully provides some clarification.

[Observation: If a teacher is good and knows his stuff, and you ask him to explain again, he would give you an alternative explanation or two to help you understand. If a teacher is not in mastery of a topic, it is easy enough for him to provide an example or solution taken from another source but which he cannot solve himself I've sometimes heard students complaining that when asked to explain again when they still cannot understand, the teacher repeats the explanation almost word for word; this usually means the teacher did not do the solution himself and would be frustrated if you keep badgering him to explain.]

Drongosky, thanks for your help but my real question is whether you sketch half of the hyperbola or the WHOLE hyperbola lets say x^2/6^2 - y^2/8^2 = 1. My teacher came up with these examples because they weren't in any textbook he looked through (havent checked Corroneos cos neither he nor I use it).

 

[Drongoski]

In my view, you should show the whole hyperbola - both halves, not in great accuracy or detail; just a sketch with the key features like the 2 foci and a typical point P(x,y) joining the two foci and the 2 vertices (-a,0) and (a,0).

 

[clintmyster]

In my view, you should show the whole hyperbola - both halves, not in great accuracy or detail; just a sketch with the key features like the 2 foci and a typical point P(x,y) joining the two foci and the 2 vertices (-a,0) and (a,0).

thats the same view i have so thats good. Anyone else?

 

[Timothy.Siu]

why would u just do half?

 

[clintmyster]

why would u just do half?

I have no idea thats why I thought my teacher was wrong (and he hardly ever is) so I came here to hopefully find no-one has heard of sketching half the hyperbola.