TANGENT AT A GIVEN POINT ON THE STANDARD PARABOLA
The standard parabola is represented by the equation
Let P1 with coordinates (x1,y1) be a point on the curve. Choose P' on the curve, figure 3-4, near the given point so that the coordinates of P' are (x2,y2). As previously stated
so that by rearranging terms, the coordinates of P' may be written as
Since P' is a point on the curve,
y2 = 4ax
the values of its coordinates may be substituted for x and y. This gives
The point P1(x1,y1) 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
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
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.
The quantity is the slope of the line connecting P1 and P. From figure 3-4, the slope of the curve at P, is obviously different from the slope of the line connecting P1 and P'.
As x and y approach zero, the ratio will approach more and more closely the true slope of the curve at P1. We designate the slope by m. Thus, as x approaches zero, equation (3.4) becomes
The equation for a straight line in the point-slope form is
Substituting for m gives
Clearing fractions, we have
Adding equations (3.5) and (3.6) yields
Dividing by y1 gives
which is an equation of a straight line in the slope-intercept form. This is the equation of the tangent line to the parabola
at the point (x1,y1).
EXAMPLE: Given the equation
find the slope of the curve and the equation of the tangent line at the point (2,4).
has the form
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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