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You need more than one tangent to find the equation of a parabola.
a parabola doesn't have one slope, the slope is constantly changing as you move accross the graph. however, it is possible to find a slope to a line tangent to a point on a parabola. to do this, take the derivative of the equation for the given parabola. then, take the X,Y coordinate and plug in the x value for the point. So, if the graph of of the equation was given by y=x^2, the derivative would be dy/dx=2x. you would then take a point, e.x. (2,4) and plug in the x value, 2, into dy/dx=2x, yielding a slope of 4 for the line tangent to that point.
A cubic line is in a cube shape line. and Parabola is a straight line
Such a line is called a tangent line or a tangent to the circle. [Tangent is Latin for touching-- a tangent line touches the circle at just one point. ]
A tangent is a line which touches, but does not cross, a curved line.
Tangent line is a graph. This graph is to gather data.
A straight line touching a circle is called a tangent. The following is the image of a tangent to a circle with center C and radius AC. The tangent touches the circle at only one point - A. visit our page: balajidentalhospital .com
The Tangent Line to Circle Theorem states that a line is tangent to a circle if and only if it's perpendicular to the circle's radius.
A common tangent is a line which is tangent to two (or more) curves.
There are several ways of defining a parabola. Here are some:Given a straight line and a point not on that line, a parabola is the locus of all points that are equidistant from that point (the focus) and the line (directrix).A parabola is the intersection of the surface of a right circular cone and a plane parallel to a generating line of that surface.A parabola is the graph of a quadratic equation.
A line, although often a segment is also called a tangent.
First we need to find the equation of the tangent line to the parabola at (2, 20).Step 1. Take the derivative of the function of the parabola.Let f(x) = 5x^2f'(x) = 10xStep 2. Find the slope of the tangent line at x = 2. Evaluate f'(2).f'(2) = 2 x 10 = 20Step 3. Using the slope, m = 20, and the point (2, 20), find the equation of the tangent line at that point. Use the point-slope form of a line(y - y1) = m(x - x1)(y - 20) = 20(x - 2)y - 20 = 20x - 40 add 20 to both sidesy = 20x - 20Step 4. Find the points of intersections of y = 5x^2 and y = 20x - 205x^2 = 20x - 20 Divide by 5 to both sidesx^2 = 4x - 4 subtract 4x and add 4 to both sidesx^2 - 4x + 4 = 0 factor(x - 2)^2= 0x = 2Step 5. Find the intersection of the tangent line with x-axis.y = 20x - 20y = 020x - 20 = 0x = 1Since the vertex of the parabola is (0, 0) and the intersection of the tangent line with parabola is (2,20) we use the interval [0, 2] to fin the required area.Step 6. IntegrateA = ∫ [(5x^2)] dx, where the below boundary is 0, and the upper boundary is 2 minus A= ∫ (20x + 20)] dx from 1 to 2= 10/3