Answer to Question #818 in Complex Analysis for victor
(a) Show that the line y = x + 2 is a tangent to the parabola y = x[sup]2[/sup]- 5x + 11.
(b) Let Q be the point where these two functions intersect. Determine the equation of thenormal to the parabola at Q.
(c) Find the area of the region enclosed between the parabola and the normal to theparabola at Q.
Let's find the points of intersection of two graphs: x2 - 5x +11 = x + 2;x2 - 6x + 9 = 0;D = 9 - 9 = 0x = 3. f(x) = 5;Thus we have only the point of intersection (3,5) - it already means that y = x + 2 is tangent to the parabola.The equation of the tangent line to the parabola f (x) - f(x0) = f'(x0)(x-x0) . f'(x0) = 2x0 - 5 = 2*3 - 5 = 1;f(x) - 5 =1 (x - x0) = 1(x - 3) = x - 3; f(x) = x +2.The equation of the normal through the point (3,5) in the form y = mx +b : m = - 1/f'(x0) = -1.5 = -1* (3) + b: b = 8;Thus the equation to the parabola would be y = -x + 8