A curve has parametric equations $x = an^2 t, \quad y = \\sin t, \quad 0 < t < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 8

First, we find the coordinates of the point at (t = \frac{\pi}{4}):

• (x = \tan^2\left(\frac{\pi}{4}\right) = 1)
• (y = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2})

Now, we calculate (\frac{dy}{dx}) at this point:

• Using (t = \frac{\pi}{4}) in the expression:

$\frac{dy}{dx} = \frac{\cos^3\left(\frac{\pi}{4}\right)}{2\sin\left(\frac{\pi}{4}\right)} = \frac{\left(\frac{\sqrt{2}}{2}\right)^3}{2\cdot\frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{4}}{\sqrt{2}} = \frac{1}{4}$

Thus, the slope of the tangent line (m = \frac{1}{4}).

Using point-slope form, we have:

$y - \frac{\sqrt{2}}{2} = \frac{1}{4}(x - 1)$

Simplifying gives:

$y = \frac{1}{4}x + \left(\frac{\sqrt{2}}{2} - \frac{1}{4}\right)$

Starting from the parametric equations:

• We have (x = \tan^2 t) and (y = \sin t).

Using the Pythagorean identity: $\sin^2 t + \tan^2 t = 1$

Substituting (x = \tan^2 t) into the equation gives:

• Rearranging leads to:

$y^2 = 1 - x$

Thus, the Cartesian equation is: $y^2 = 1 - x$