4-10.2
Find the largest and smallest distances from the origin to the
conic whose equation is 5x2 − 6xy + 5y 2 − 32 = 0 and hence determine
the lengths of the semiaxes of this conic.
Considering the square of the distance as usual:
L = x2 + y 2 + λ(5x2 − 6xy + 5y 2 )
Then
Lx = 2x + λ(10x − 6y) = 0
Ly = 2y + λ(−6x + 10y) = 0
Solving (multiply 1st by y, 2nd by x, and subtract): λ(6x2 − 6y 2 ) = 0 ⇒
Cases : λ = 0 (which implies x = y = 0 (that is not on the conic)
or x2 = y 2
√
Two cases: x = √
y ⇒ 4x2 = 32 ⇒ |x| = 2 2 = |y| or√x =√ −y ⇒
16x2√= 32 √
⇒ |x| = 2 = |y|. Thus there
√ are
√ four CPs,
√ (2√ 2, 2 2) and
(−2 2, −2 2) which are maxima and ( 2, − 2) and (− 2, 2) which are
minima. The semiaxes are 4 and 2 for this ellipse (see the picture).
1
Since B/(A − C) = −6/(5 − 5) = ∞, and arctan ∞ =√π/2, rotate the
axes by π/4 as follows (with sin(π/4) = cos(π/4) = 1/ 2):
x = p cos θ − q sin θ
y = p sin θ + q cos θ
q
p
5 √ −√
2
2
!2
!
!
p
q
p
q
p
q
−6 √ − √ × √ + √ +5 √ + √
2
2
2
2
2
2
!2
5
5 2
(p − 2pq + q 2 ) − 3(p2 − q 2 ) + (p2 + 2pq + q 2 )
2
2
2
2
2p + 8q = 32
p2 q 2
+
=1
42 22
That is, an ellipse with semiaxes 4 and 2 along the 45 degree lines. In
the original coordinate system, the endpoints are then as before.
2