Suppose for example an ellipse $x^2-2xy+2y^2=2$. How can we find the equation of the line through its major axis?
Another question on its area is possibly easy.
Asked
Active
Viewed 287 times
0
-
Consider rotation of axes – Shailesh Sep 30 '15 at 05:36
-
@Shailesh You will find it is quite hard to orthogonalize $A$. – xldd Sep 30 '15 at 05:44
-
See if this helps or our own SE site see this – Shailesh Sep 30 '15 at 06:05
-
This is the equation rotated through $\frac{\pi}{8}$ – Shailesh Sep 30 '15 at 06:16
2 Answers
1
The ellipse is obviously centered at the origin, and the axis are of the form $y=mx$.
Substituting in the ellipse equation,
$$x^2-2mx^2+2m^2x^2=2$$ you get the intersection points
$$(x,y)=\pm\left(\frac2{\sqrt{1-2m+2m^2}},\frac{2m}{\sqrt{1-2m+2m^2}}\right).$$
The squared distance to the center is $$4\frac{1+m^2}{1-2m+2m^2}.$$
By cancelling the derivative, you find two solutions,
$$-\phi,\phi^{-1}$$ where $\phi$ is the Golden section. (Note that the product is $-1$, as should be.)
The corresponding squared distances (i.e. semi-axis lengths) are $$4\phi^{-2},4\phi^2.$$
-
Caution: this is a do-it-yourself solution, probably not the one expected by the teacher. – Sep 30 '15 at 06:49
0
Change to orthogonal co-ordinate axes whose equations in $x,y$ are $y=x\tan W$ and $y=x\cot W$ with $W$ to be determined.So $x=u\cos W + v\in W$ and $y=-u\sin W +v \cos W$ . Substitute into the equation for the ellipse and put the co-efficient of $u v$ to be $0$. This imposes a condition of the form $ 2 A\sin W \cos W =B(\cos^2 W-\sin^2 W)$, that is, $A\sin 2 W=B \cos 2 W$. Solve for $W$. The major axis of the ellipse, in terms of $x,y$ is on the line $y=x \tan W$.The equation of the ellipse ,in terms of $u,v$ is of the form $Cu^2+Dv^2=E$ with positive $C.D.E$, from which the area is readily found.
DanielWainfleet
- 57,985