A curve has parametric equations x = tan²(t), y = sin(t), 0 < t < π/2 - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 7

First, we need to find the coordinates at ( t = \frac{\pi}{4} ).

1. Calculate ( x ) and ( y ):
   • ( x = \tan²\left(\frac{\pi}{4}\right) = 1 )
   • ( y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} )

Thus, the point is ( (1, \frac{\sqrt{2}}{2}) ).

2. Now, find ( \frac{dy}{dx} ) at ( t = \frac{\pi}{4} ):

   • ( \frac{dx}{dt} = 2\tan\left(\frac{\pi}{4}\right)\sec²\left(\frac{\pi}{4}\right) = 2 )
   • ( \frac{dy}{dt} = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} )
   • Therefore, ( \frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2}}{2} = \frac{\sqrt{2}}{4} )

3. The equation of the tangent line:

   • We can use point-slope form: ( y - y_1 = m(x - x_1) )
   • This gives us: ( y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}(x - 1) )
   • Rearranging gives: ( y = \frac{\sqrt{2}}{4}x + \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{4}\right) )

Starting from the parametric equations, we have:

1. From ( x = \tan²(t) ), we find ( \tan(t) = \sqrt{x} ).

2. We also know that ( \sin^2(t) + \cos^2(t) = 1 ).

3. Using this identity: [ \sin^2(t) = 1 - \cos^2(t) \] But since ( \tan(t) = \frac{\sin(t)}{\cos(t)} ), we can write ( \cos(t) = \frac{\sin(t)}{\tan(t)} = \frac{\sin(t)}{\sqrt{x}} ).

4. Substitute back to express in terms of ( x ):

   • This leads to the relation between ( y^2 ) and ( x ): [ y^2 = 1 - \frac{1}{x + 1} \text{ (after substituting and rearranging)} ]