Show that the tangent at P(ct, c/t) on the curve x=ct, y=c/t has equation x+t2y=2ct. Show the area of triangle QOR is constant.

You're on a roll — Let's solve questions 7 and 8 from the image you sent. Questions 6 i) and 6 ii) were previously addressed (they were numbered as 5 i) and 5 ii) in the original context). --- Question 7: Show that the tangent at the point P, with parameter t, on the curve whose parametric equations are x=ct, y=(c)/(t), where c is a constant, has equation x+t^2y = 2ct. This tangent meets the x-axis at the point Q and the line through P parallel to the x-axis cuts the y-axis at a point R. Show that the area of triangle QOR, where O is the origin, is a constant. Step 1: Find the derivatives (dx)/(dt) and (dy)/(dt). Given x = ct, so (dx)/(dt) = c. Given y = (c)/(t) = ct^-1, so (dy)/(dt) = -ct^-2 = -(c)/(t^2). Step 2: Find the gradient of the tangent (dy)/(dx). (dy)/(dx) = (dy/dt)/(dx/dt) = (-c/t^2)/(c) = -(1)/(t^2) Step 3: Find the equation of the tangent at point P. The coordinates of point P are (x_P, y_P) = (ct, (c)/(t)). Using the point-slope form y - y_P = m(x - x_P): y - (c)/(t) = -(1)/(t^2)(x - ct) Multiply the entire equation by t^2 to clear denominators: t^2y - t^2((c)/(t)) = -(x - ct) t^2y - ct = -x + ct Rearrange the terms to match the required form: x + t^2y = ct + ct x + t^2y = 2ct This shows the equation of the tangent. Step 4: Find the coordinates of point Q. Point Q is where the tangent meets the x-axis, so y=0. Substitute y=0 into the tangent equation x+t^2y = 2ct: x + t^2(0) = 2ct x = 2ct So, point Q is (2ct, 0). Step 5: Find the coordinates of point R. Point R is where the line through P parallel to the x-axis cuts the y-axis. A line through P parallel to the x-axis has the equation y = y_P. Since P = (ct, (c)/(t)), the equation of this line is y = (c)/(t). This line cuts the y-axis when x=0. So, point R is (0, (c)/(t)). Step 6: Show that the area of triangle QOR is a constant. The vertices of the triangle are O(0,0), Q(2ct, 0), and R(0, (c)/(t)). This is a right-angled triangle with legs along the x and y axes. The length of the base OQ is |2ct|. The length of the height OR is |(c)/(t)|. The area of triangle QOR is given by: Area = (1)/(2) × base × height Assuming c>0 and t>0 (or considering absolute values for general c, t): Area = (1)/(2) × (2ct) × ((c)/(t)) Area = (1)/(2) × 2c^2 Area = c^2 Since c is a constant, c^2 is also a constant. The area of triangle QOR is c^2. --- Question 8: Express (2)/((1+x)(1+3x)) in partial fractions. Hence solve the differential equation $\