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

1. First, find the coordinates at $t = \frac{\pi}{4}$: $x = \tan^2\left(\frac{\pi}{4}\right) = 1, \quad y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.$
   The point is $(1, \frac{\sqrt{2}}{2}).$

2. Now find $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$: $\frac{dy}{dx} = \frac{\cos^3(\frac{\pi}{4})}{2\sin(\frac{\pi}{4})} = \frac{(\frac{\sqrt{2}}{2})^3}{2(\frac{\sqrt{2}}{2})} = \frac{\sqrt{2}}{2}.$

3. Use point-slope form of the equation of the tangent: $y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(x - 1).$
   Rearranging gives: $$y = \frac{\sqrt{2}}{2}x + \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2}.$

From the parametric equations, we know:

1. $x = \tan^2 t$ implies $\tan t = \sqrt{x}$.

2. Substituting $\tan t$ into $y = \sin t$: $y = \sin t = \frac{\tan t}{\sqrt{1 + \tan^2 t}} = \frac{\sqrt{x}}{\sqrt{1 + x}}.$

3. Therefore, squaring both sides hence gives: $y^2 = \frac{x}{1 + x}.$
   Thus, the cartesian equation is: $y^2 = \frac{x}{1 + x}.$