A curve has parametric equations $x = an^2 t, \\ y = ext{sin} t, \\ 0 < t < \frac{\pi}{2}.$ (a) Find an expression for \( \frac{dy}{dx} \) in terms of \( t \) - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 8

At ( t = \frac{\pi}{4} ), calculate:

1. The coordinates of the point:

   • ( x = \tan^2(\frac{\pi}{4}) = 1 ) and ( y = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} ).
   • Therefore, the point is ( (1, \frac{\sqrt{2}}{2}) ).

2. The slope of the tangent:

   • Evaluate ( \frac{dy}{dx} ) at ( t = \frac{\pi}{4} ).
   • ( \frac{dy}{dx} = \frac{\cos^3(\frac{\pi}{4})}{2\sin(\frac{\pi}{4})} = \frac{(\frac{\sqrt{2}}{2})^3}{2 \cdot \frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{4}}{\sqrt{2}} = \frac{1}{4} ).

3. Use point-slope form to find the equation of the tangent line:

   • ( y - y_1 = m(x - x_1) ), which gives:
   • ( y - \frac{\sqrt{2}}{2} = \frac{1}{4}(x - 1) ).
   • Rearranging yields the equation in the form ( y = ax + b ) with ( a = \frac{1}{4} ) and ( b = \frac{\sqrt{2}}{2} - \frac{1}{4} ).

To eliminate ( t ):

1. From ( y = \sin t ) we have ( t = \arcsin(y) ).

2. Substitute into the ( x ) equation:

   • ( x = \tan^2(t) = \tan^2(\arcsin(y)) = \frac{y^2}{1 - y^2} ).

3. Rearranging gives:

   • ( y^2 = x(1 - y^2) ), leading to the form ( y^2 = \frac{x}{1 + x} ).