A hyperbola locus problem
H:
Let the left and right vertices of the hyperbola
x2
3
− y 2 = 1 be A1 , A2 .
Let P(x1 , y1 ), Q(x1 , −y1 ) be two different movable points on the given
hyperbola.
Let the intersection of A1 P and A2 Q be M. Find the equation E of the
locus of M.
Method 1
From the question
|x1 | > 3 and
A1 (−√3, 0), A2 (√3, 0) .
y1
PA1 : y =
(x + √3) … . (1)
x1 + √3
−y1
QA2 : y =
(x − √3) … . (2)
x1 − √3
3
√3y1
x1
Solving, we get:
x=x , y=
Change of subjects, we get:
x1 = x , y1 =
Then
1
3
√3y
x
… . (3)
x ≠ 0, |x| < 3 .
Since P(x1 , y1 ) is on the hyperbola :
x2
(3) ↓ (4), E:
3
− y 2 = 1 , we have
x2
+ y 2 = 1, x ≠ 0,
3
x1 2
3
− y1 2 = 1 … . (4)
x ≠ ±√3
1
In the above, x ≠ 0, x ≠ ±√3 should be included in E. We can also get this by another way.
Assertion: x ≠ 0, x ≠ ±√3 in E.
P, Q are two different points on the given hyperbola, so they cannot coincide with A1 and A2 .
Therefore A1 and A2 cannot be points in E and x ≠ ±√3
Let L be the line joining (0,1) and A2 (√3, 0) .
Solving :
{
L: x + √3y − √3 = 0
H:
x2
3
− y2 = 1
We get x = √3, y = 0 , Hence the only intersection point between L and H is A2 (√3, 0).
Hence E cannot pass through (0,1) .
Similarly E cannot pass through (0, −1) .
Method 2
From the question |x1 | > 3 and A1 (−√3, 0), A2 (√3, 0) .
Let M(x, y) be the intersection point of A1 P and A2 Q .
y
A1 , M, P are collinear,
x+√3
A2 , M, Q are collinear,
x−√3
(5) × (6),
y2
y
−y1 2
x2 −3
=x
1
2 −3
=x
y1
1 +√3
=x
−y1
1 −√3
x1 2
3
… . (6)
… . (7)
Since P(x1 , y1 ) is on the hyperbola :
−y1 2 = 1 −
… . (5)
x2
3
− y 2 = 1 , we have
x1 2
3
− y1 2 = 1.
… . (8)
x2
(8) ↓ (7), E:
+ y2 = 1
3
By the Assertion , x ≠ 0, x ≠ ±√3 .
Further point of interest for you to try :
Let the left and right vertices of the ellipse
E:
x2
3
− y 2 = 1 be A1 , A2 .
Let P(x1 , y1 ), Q(x1 , −y1 ) be two different movable points on the given ellipse.
Let the intersection of A1 P and A2 Q be M. Find the equation H of the locus of M.
2