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TANGENT AT A GIVEN POINT ON THE STANDARD PARABOLA The standard parabola is represented by the equation
Let P_{1} with coordinates (x_{1},y_{1})
be a point on the curve. Choose P' on the curve, figure 34, near the given point so that the coordinates of P' are (x_{2},y_{2}). As
previously stated
Figure 34.Parabola.
and
so that by rearranging terms, the coordinates of P' may be
written as
Since P' is a point on the curve, y^{2} = 4ax the values of its coordinates may be substituted for x and
y. This gives
or
The point P_{1}(x_{1},y_{1}) also
lies on the curve, so we have
Substituting this value for y; into equation (3.1)
transforms it into
Simplifying, we obtain
Divide both sides by Ax, obtaining
which gives _{
} Solving for
we find
Before proceeding, we need to discuss the term
in equation (3.3). If we solve equation (3.2) for
y, we find
then
and
Since the denominator contains a term not dependent upon
y or
x, as we
let
x approach
zero,
y will
also approach zero. NOTE: We may find a value for Ax that will make
y less
than 1; then when
y is
squared, it will approach zero at least as rapidly as
x does. We now refer to equation (3.3) again and make the
statement so that we may disregard
, since it approaches zero when Ax approaches zero. Then
The quantity
is the slope of the line connecting P_{1}
and P. From figure 34, the slope of the curve at P, is obviously different
from the slope of the line connecting P_{1} and P'. As
x and
y approach
zero, the ratio
will
approach more and more closely the true slope of the curve at P_{1}. We
designate the slope by m. Thus, as
x approaches zero, equation (3.4) becomes
The equation for a straight line in the pointslope form
is
Substituting
for m gives
Clearing fractions, we have
but
Adding equations (3.5) and (3.6) yields
Dividing by y_{1} gives
which is an equation of a straight line in the
slopeintercept form. This is the equation of the tangent line to the parabola
at the point (x_{1},y_{1}). EXAMPLE: Given the equation
find the slope of the curve and the equation of the
tangent line at the point (2,4). SOLUTION:
has the form
where
and 2a=4 The slope, m, at point (2,4) becomes
Since the slope of the line is 1, then the equation of the
tangent to the curve at the point (2,4) is

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