Ellipse Question (1 Viewer)

[George W. Bush]

If E(1) = x^2 + 3y^2 - 1 = 0, E(2) = 4x^2 + y^2 - 1

a) If E(1) = 0 and E(2) = 0, show that k is a real number where E(1) + kE(2) = 0 is a curve passing through the intersections of E(1) and E(2)
b) Find all values of k for which E(1) + kE(2) = 0 is an ellipse.

Fully justify please (you can skip the totally easy steps)

 

[maniacguy]

I'm going to say what I understand your question to mean, because I'm not sure exactly what you do mean:

Let E_1 := x^2 + 3y^2 - 1
Let E_2 := 4x^2 + y^2 - 1

a) Show that if k is a real number, then the curve with equation described by:
E_1 + k*E_2 = 0 passes through all intersections of the curves E_1 = 0 and E_2 = 0.

(the answer in this event is simply to let A be a point in the intersection. Then at A E_1 = 0 and E_2 = 0, so E_1+k*E_2 = 0 + k*0 = 0, i.e. A lies on the curve E_1 + k*E_2 = 0)

b) Find all k such that the curve described by E_1 + k*E_2 = 0 is an ellipse.

In this case,
E_1 + k*E_2
= (x^2+3y^2-1) + k*(4x^2+y^2-1)
= (1+4k)x^2 + (3+k)y^2 - (1+k)

so E_1 + k*E_2 = 0 is equivalent to:
(1+4k)x^2 + (3+k)y^2 = (1+k)

To put this in the standard form for an ellipse (i.e. x^2/a^2 + y^2/b^2 = 1), we need 1+k not equal to 0, i.e. k<>-1.
(Otherwise you don't get an ellipse, you instead get the union of two lines intersecting at the origin).

We also need two other things:
i) (4k+1)/(k+1) > 0, so k<-1 or k>-1/4
ii) (3+k)/(k+1) > 0, so k< -3 or k > -1

These ensure that we do get a^2 and b^2 - i.e. the squares are positive

Thus the values of k that generate ellipses are k<-3 and k>-1/4

 

[maniacguy]

(For anyone who's interested, try visualizing what happens as well for the different values of k. It's quite interesting.)

 

[George W. Bush]

(1+4k)x^2 + (3+k)y^2 = (1+k)

At this line, for this to be an ellipse do we also need

(1+4k)(3+k) = (1+k)
as an ellipse is of the form x^2b^2 + y^2a^2 =a^2b^2?

 

[maniacguy]

No - an ellipse is of the form x^2b^2 + y^2a^2 = a^2b^2, yes.

It is also of the form
x^2*(C*b^2) + y^2*(C*a^2) = C*a^2b^2

Thus:
(1+4k)(1+3k) = C*b^2 * C*a^2
= C*(Ca^2*b^2)
= C*(1+k)

Incidentally, where did the original question come from?

 

[George W. Bush]

Half-yearly

 

[sammeh]

correct me if im wrong (i probly am - have had glandular fever and missed 5 weeks of school including the conics work, so today im relearning for half yearlies next week :S) but k would be an expression for the eccentricity?

 

[CM_Tutor]

Originally posted by sammeh
correct me if im wrong (i probly am - have had glandular fever and missed 5 weeks of school including the conics work, so today im relearning for half yearlies next week :S) but k would be an expression for the eccentricity?

No, k is not the eccentricity, as k can be negative, and the eccentricity cannot. However, it would be possible to find an expression for the eccentricity in terms of k.

 

[sammeh]

well thats pretty much what i meant

 

[CM_Tutor]

Originally posted by sammeh
well thats pretty much what i meant

OK, then I modify my answer:

Yes, the eccentricity can be expressed in terms of k.

 

[George W. Bush]

You'd also have to consider a vertical ellipse though, am I right?