AQA PC4 2015 Q5 // A) Find the gradient at P. B) Find the equation of the normal to the curve at P C)The normal P intersects at the curve again at the point Q(cos2q, sin q) Hence find the x-coordinate of Q.
A) We need to find dy/dx. We know that dy/dx =(dy/dt)/(dx/dt). Differentiating each parametric equation we get: dx/dt= -2sin(2t) {using the chain rule and knowing that cos differentiates to -sin.} dy/dt= cos(t) {knowing that sin differentiates to cos.} Next we combine these two expressions in order to get dy/dx = cos(t)/-2sin(2t). Thus substituting when t = Pi/6 we get the gradient at P = -1/2. B) The gradient of a normal is a negative reciprocal of the tangent at P - therefore the gradient of P is 2. This gives us a starting equation of y = 2x + c. By substituting t = Pi/6 into our two parametric equations for x and y we get x = 1/2 and y = 1/2. This is then subistituted into the equation y = 2x + c and rearranged to get c = -1/2. The final answer should be given as y = 2x - 1/2. C) To start this part we need to find an way of expressing the parametric equation non-parametrically ie only one variable. From the trigonometric identities we know that cos 2t = 1-2sin2t. Because y = sin t, this leaves us with the expression x = 1-2y2. Now the normal can be rearranged to the form x = (y+1/2)/2. Equating both expressions for x we get 1-2y2 = (y+1/2)/2. This can be rearranged into the quadratic 8y2+2y-3 = 0. The point Q has y co-ordinate sin q. Subbing into our quadratic equation and solving we get sin q = 1/2 and sin q = -3/4. We can ignore the sin q = 1/2 as this is the point P intersect. Next to find the x-coordinate of Q we can substitute y = -3/4 (as y = sin q) into the equation of the normal. Solving for x this gives x = -1/8. This is the x-coordinate of Q.
