MARK SCHEME – A-LEVEL MATHEMATICS – 7357/2 – JUNE 202012QMarking instructionsAOMarkTypical solution8(a)Eliminates t1.1aM1𝑦𝑦2=𝑡𝑡,𝑥𝑥=𝑦𝑦²4𝑦𝑦² = 4𝑥𝑥Writes the Cartesian equationin the required form1.1bA1Subtotal28(b)(i)Differentiates both𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡= 2𝑡𝑡,𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡= 2with at least one correctOrDifferentiates their𝑦𝑦² = 4𝑥𝑥toobtainddyyAx=Or rearranges and differentiates2yx=and obtains12ddyAxx−=OE3.1aM1𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡= 2𝑡𝑡,𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡= 2𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥=22𝑎𝑎=1𝑎𝑎The gradient of a line is equal tothe tangent of the angle betweenthe line and the horizontal hence1tanaθ=Obtains correct𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥att=a1.1bA1Explains that the gradient of aline is the tangent of the anglebetween the line and thehorizontalorshows on right-angled triangleon diagram and links totanθandconcludes1tanaθ=2.4E1Subtotal38(b)(ii)Uses formula for gradient ofstraight line with points A and BMust have21a−or21a−indenominator1.1aM12220tan121aaaaφ−=−=−Obtains correcttanφOE1.1bA1Subtotal2