A-Level Differentiation: worked solution

Question

With p = pi/3, the tangent Mp and normal Np meet the x-axis at Q and R. Find the area of triangle PQR.

Step 1: find the point P at p = pi/3. Substituting p = pi/3 into the coordinates of P from part (a) gives P(3/2, sqrt3/2). Step 2: find Q, where the tangent Mp meets the x-axis. At p = pi/3 the gradient of the tangent is -sqrt3, so the tangent through P is y - sqrt3/2 = -sqrt3 (x - 3/2). The x-axis is the line y = 0, so put y = 0: 0 - sqrt3/2 = -sqrt3 (x - 3/2) Divide both sides by -sqrt3: 1/2 = x - 3/2 x = 1/2 + 3/2 = 2. So Q(2, 0). Step 3: find R, where the normal Np meets the x-axis. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of -sqrt3, namely 1/sqrt3. The normal through P is y - sqrt3/2 = (1/sqrt3)(x - 3/2). Put y = 0: -sqrt3/2 = (1/sqrt3)(x - 3/2) Multiply both sides by sqrt3: -3/2 = x - 3/2 x = 0. So R(0, 0), i.e. the normal passes through the origin. Step 4: find the area of triangle PQR. Q and R both lie on the x-axis, so QR is a horizontal base: base QR = |2 - 0| = 2. The perpendicular height from P down to the x-axis is the y-coordinate of P: height = sqrt3/2. Area = (1/2) x base x height = (1/2)(2)(sqrt3/2) = sqrt3/2 square units. Check: the angle at P is a right angle (tangent perpendicular to normal), with PQ = sqrt((2 - 3/2)² + (0 - sqrt3/2)²) = sqrt(1/4 + 3/4) = 1, PR = sqrt((0 - 3/2)² + (0 - sqrt3/2)²) = sqrt(9/4 + 3/4) = sqrt3, so the area is also (1/2)(1)(sqrt3) = sqrt3/2. The two methods agree.