Assoc. Prof. Cornel Pintea Geometry for First Year students Tutorial

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Assoc. Prof. Cornel Pintea
Geometry for First Year students
Tutorial 11
1. Show that a ray of light through a focus of an ellipse reflects to a ray that
passes through the other focus (the optical property of the ellipse).
x2 y 2
Solution. Let F1 (−c, 0), F2 (c, 0) be the foci of the ellipse E : 2 + 2 =
a
b
1. Recall that the gradient grad(f )(x0 , y0 ) = (fx (x0 , y0 ), fy (x0 , y0 )) is a
normal vector of the ellipse E to its point M0 (x0 , y0 ), where
f (x, y) = δ(F1 , M ) + δ(F2 , M ) =
p
p
(x − c)2 + y 2 + (x + c)2 + y 2
and M (x, y), as the ellipse is a level set of f . Note that
fx (x0 , y0 ) =
x0 − c
x0 + c
y
y
+
and fy (x0 , y0 ) =
+
,
δ(F1 , M0 ) δ(F2 , M0 )
δ(F1 , M0 ) δ(F2 , M0 )
and shows that
grad(f ) = (fx (x0 , y0 ), fy (x0 , y0 ))
x0 − c
x0 + c
y0
y0
=
+
,
+
δ(F1 , M0 ) δ(F2 , M0 ) δ(F1 , M0 ) δ(F2 , M0 )
−→
−→
(x0 − c, y) (x0 + c, y)
F1 M 0
F2 M0
=
+
=
+
.
δ(F1 , M0 )
δ(F2 , M0 )
δ(F1 , M0 ) δ(F2 , M0 )
−→
−→
F1 M 0
F2 M 0
The versors
and
point towards the exterior of the
δ(F1 , M0 )
δ(F2 , M0 )
ellipse E and their sum make obviously equal angles with the directions
−→
−→
of the vectors F1 M0 and F2 M0 and (the sum) is also orthogonal to the
tangent TM0 (E) of the ellipse at M0 (x0 , y0 ). This shows that the angle
between the ray F1 M and the tangent TM0 (E) equals the angle between
the ray F2 M and the tangent TM0 (E).
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2. Show that a ray of light through a focus of a hyperbola reflects to a ray
that passes through the other focus (the optical property of the hyperbola).
x2 y 2
−
=
a2 b2
1. Recall that the gradient grad(f )(x0 , y0 ) = (fx (x0 , y0 ), fy (x0 , y0 )) is a
normal vector of the hyperbola H to its point M0 (x0 , y0 ), where
Solution. Let F1 (−c, 0), F2 (c, 0) be the foci of the hyperbola E :
p
(x − c)2 + y 2 −
p
(x + c)2 + y 2
(1)
on the left hand side branch of H and
p
p
f (x, y) = δ(F1 , M ) − δ(F2 , M ) = (x + c)2 + y 2 − (x − c)2 + y 2
(2)
f (x, y) = δ(F2 , M ) − δ(F1 , M ) =
on the right hand side branch of H and M (x, y). We shall only use the
version (2) of f , as the judgement for the version (1) works in a similar
way. Note that
fx (x0 , y0 ) =
x0 + c
x0 − c
y
y
−
and fy (x0 , y0 ) =
−
,
δ(F1 , M0 ) δ(F2 , M0 )
δ(F1 , M0 ) δ(F2 , M0 )
and shows that
grad(f ) = (fx (x0 , y0 ), fy (x0 , y0 ))
x0 + c
x0 − c
y0
y0
=
−
,
−
δ(F1 , M0 ) δ(F2 , M0 ) δ(F1 , M0 ) δ(F2 , M0 )
−→
−→
(x0 + c, y) (x0 − c, y)
F1 M 0
F2 M0
=
−
=
−
.
δ(F1 , M0 )
δ(F2 , M0 )
δ(F1 , M0 ) δ(F2 , M0 )
3
−→
−→
F1 M0
F2 M0
The versors
and −
point towards the ’exterior’ of the
δ(F1 , M0 )
δ(F2 , M0 )
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hyperbola H and their sum make obviously equal angles with the direc−→
−→
tions of the vectors F1 M0 and F2 M0 and (the sum) is also orthogonal to
the tangent TM0 (H) of the hyperbola at M0 (x0 , y0 ), as it is the gradient of
f and the gradient is normal to the right hand side branch of the hyperbola. In deed the right hand side of the hiperbola is a level set of f . This
shows that the angle between the ray F1 M and the tangent TM0 (H) equals
the angle between the ray F2 M and the tangent TM0 (H).
3. Show that a ray of light through a focus of a parabola reflects to a ray
parallel to the axis of the parabola (the optical property of the parabola).
Solution. Let F ( p2 , 0) be the focus of the parabola P : y 2 = 2px and
d : x = − p2 be its director line. Recall that the gradient grad(f )(x0 , y0 ) =
(fx (x0 , y0 ), fy (x0 , y0 )) is a normal vector of parabola P to its point M0 (x0 , y0 ),
where
f (x, y) = δ(F, M ) − δ(M, d) =
r
x−
p
p 2
+ y2 − x +
2
2
and M (x, y). Note that
fx (x0 , y0 ) =
1
x0 − p2
y0
− 1 and fy (x0 , y0 ) =
,
δ(F, M0 )
δ(F, M0 )
The exterior of a hyperbola is the nonconvex component of its complement
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and shows that
grad(f ) = (fx (x0 , y0 ), fy (x0 , y0 )) x0 − p2
y0
=
− 1,
δ(F, M0 )
δ(F, M0 )
−→
x0 − p2
y0
F M0
=
,
− (1, 0) =
− i.
δ(F, M0 ) δ(F, M0 )
δ(F, M0 )
−→
F M0
The versors
and −i point towards the ’exterior’ of the parabola
δ(F, M0 )
P 2 and their sum make obviously equal angles with the directions of the
−→
vectors F M0 and i and (the sum) is also orthogonal to the tangent line
TM0 (P) of the parabola at M0 (x0 , y0 ). This shows that the angle between
the ray F M and the tangent line TM0 (P) equals the angle between Ox and
the tangent TM0 (E).
4. Find the rectilinear generatrices of the hyperboloid of one sheet
(H1 )
x2
y2
z2
+
−
= 1 which are parallel to the plane (π) x + y + z = 0.
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9
4
Hint. The equations of the two familes of generatrices should be considered
as well as their director parameters. The lines of each family which are parallel to the given plane should be selected out of the parallelism condition
between a line and a plane.
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The exterior of a parabola is the nonconvex component of its complement