Tangents to Curves
A review of some ideas, relevant
to the calculus, from high school
plane geometry
Straightedge and Compass
• The physical tools for drawing the figures
that Plane Geometry investigates are:
– The unmarked ruler (i.e., a ‘straightedge’)
– The compass (used for drawing of circles)
Lines and Circles
• Given any two distinct points, we can use
our straightedge to draw a unique straight
line that passes through both of the points
• Given any fixed point in the plane, and any
fixed distance, we can use our compass to
draw a unique circle having the point as its
center and the distance as its radius
The ‘perpendicular bisector’
• Given any two points P and Q, we can
draw a line through the midpoint M that
makes a right-angle with segment PQ
P
Q
M
Tangent-line to a Circle
• Given a circle, and any point on it, we can
draw a straight line through the point that
will be tangent to this circle
How do we do it?
• Step 1: Draw the line through C and T
C
T
How? (continued)
• Step 2: Draw a circle about T that passes
through C, and let D denote the other end
of that circle’s diameter
C
T
D
How? (contunued)
• Step 3: Construct the straight line which is
the perpendicular bisector of segment CD
tangent-line
C
T
D
Proof that it’s a tangent
• Any other point S on the dotted line will be
too far from C to lie on the shaded circle
(because CS is the hypotenuse of ΔCTS)
S
C
T
D
Tangent-line to a parabola
• Given a parabola, and any point on it, we
can draw a straight line through the point
that will be tangent to this parabola
axis
focus
directrix
parabola
How do we do it?
• Step 1: Drop a perpendicular from T to the
parabola’s directrix; denote its foot by A
T
A
F
focus
axis
directrix
parabola
How? (continued)
• Step 2: Locate the midpoint M of the linesegment joining A to the focus F
T
A
M
F
focus
axis
directrix
parabola
How? (continued)
• Step 3: Construct the line through M and T
(it will be the parabola’s tangent-line at T,
even if it doesn’t look like it in this picture)
T
A
tangent-line
M
F
focus
axis
directrix
parabola
Proof that it’s a tangent
• Observe that line MT is the perpendicular
bisector of segment AF (because ΔAFT
will be an isosceles triangle)
T
A
tangent-line
TF = TA
because T is
on the parabola
M
F
focus
axis
directrix
parabola
Proof (continued)
• So every other point S that lies on the line
through points M and T will not be at equal
distances from the focus and the directrix
SB < SA
B
since SA is hypotenuse
A
of right-triangle ΔSAB
SA = SF
because SA lies on AF’s
perpendictlar bisector
S
T
M
F
focus
axis
Therefore: SB < SF
directrix
parabola
Tangent to an ellipse
• Given an ellipse, and any point on it, we
can draw a straight line through the point
that will be tangent to this ellipse
F1
F2
How do we do it?
• Step 1: Draw a line through the point T
and through one of the two foci, say F1
T
F1
F2
How? (continued)
• Step 2: Draw a circle about T that passes
through F2, and let D denote the other end
of that circle’s diameter
T
F1
F2
D
How? (continued)
• Step 3: Locate the midpoint M of the linesegment joining F2 and D
T
D
M
F1
F2
How? (continued)
• Step 4: Construct the line through M and T
(it will be the ellipse’s tangent-line at T,
even if it doesn’t look like it in this picture)
T
D
M
F1
F2
tangent-line
Proof that it’s a tangent
• Observe that line MT is the perpendicular
bisector of segment DF2 (because ΔTDF2
will be an isosceles triangle)
T
D
M
F1
F2
tangent-line
Proof (continued)
• So every other point S that lies on the line
through points M and T will not obey the
ellipse requirement for sum-of-distances
S
T
D
M
F1
F2
tangent-line
SF1 + SF2 > TF1 + TF2 (because SF2 = SD and TF2 = TD )
Why are these ideas relevant?
• When we encounter some other methods
that purport to produce tangent-lines to
these curves, we will now have a reliable
way to check that they really do work!