13
My way to prove β is equal to α is to prove that the complementary angles of
β and α are equal. So two perpendiculars are drawn from the foci to the line l with the feet G and H respectively. The problem then obviously turns into proving that the two triangles are similar. Here come the details.
2
2
x y
2 A B + = ( A > B ), I can derive the coordinates of the two 2 1
From the equation
2 2 2 C A B = − , so the coordinates of foci F 1 and F 2 are ( C , 0) and
foci. Let’s say
( C − , 0) respectively.
0 y = , the tangents(on the left and right) are parallel with the
Apparently, when
y-axis. The ray will just reflect back in the same way as it comes. So it will pass through both of the points.
When y ≠ , we can get the derivative function of the ellipse by implicit 0
derivation(1).
2
2 dy B x dx A y = −
We assume the point of contact T have the coordinate ( , ) a b .
2
2 B a A b
= −
The gradient of the tangent:
gradient
So the line l , which is the tangent, has the equation 2
2
2 2
2 2
0 B ax A by A b B a + − − =
2
2
The distance between the right focus F 1 and point T: F 1 T =
( ) a C b − +
Similarly, the distance between the left focus F 2 and point T: F 2 T =
2
2
( ) a C b + +
Because we have the equation of the tangent and the coordinates of the two foci, we can calculate the distance from the foci to the line.
0 A x B y C + + = :
This is the formula. The distance from a point (x 1
y 1 ) to a line
0
0
0
0 1 0 1 + + + 2 2 A B A x B y C
0
(2)
0
0
Made with FlippingBook - Online Brochure Maker