      Custom Search   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  Figure 3-4.-Parabola.   and 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 or 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 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 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 but 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). 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    Integrated Publishing, Inc. - A (SDVOSB) Service Disabled Veteran Owned Small Business