A curve has the parametric equations
x=ln(t+1),y=t2−5,t>−1 x = \ln(t + 1), \quad y = t^2 - 5, \quad t > -1 x=ln(t+1),y=t2−5,t>−1The points where the curve crosses the coordinate axes are (ln(5+1),0)(\ln(\sqrt{5} + 1), 0)(ln(5+1),0) and (0,−5)(0, -5)(0,−5). Find the corresponding values of ttt.
The tangent to the curve has equation y=24x+4−24ln4y = 24x + 4 - 24\ln 4y=24x+4−24ln4. Find the value of t t\,t at the point of contact.
215 exam-style questions on Edexcel A Level Maths Parametric Equations, covering 8.1 Parametric Equations, 8.2 Using Trigonometric Identities, 8.3 Curve Sketching, 8.4 Points of Intersection, and 8.5 Modelling with Parametric Equations. Each one has a worked solution and a mark scheme showing where the marks go.