A curve has parametric equation $x=2t-1$ , $y=\frac{1}{{{t}^{2}}+1}$. Prove that the equation of

Question

A curve has parametric equation $x=2t-1$ , $y=\frac{1}{{{t}^{2}}+1}$. Prove that the equation of the tangent at the point with parameter $t$ is $(t^{2} + 1)^{2}y + tx = 3t^{2} - t + 1$. The tangent at the point where $t = 3$ meets the curve again at the point where $t = q$. Find the value of $q$ and hence find the coordinates of the point of intersection.

Accepted Answer

(a) $(t^2+1)^2y+tx=3t^2-t+1$ (b) $q=-\frac43$, intersection $\left(-\frac{11}{3},\frac9{25}\right)$

Techniques of Differentiation - Differentiation of Parametric Equations

Techniques of Differentiation - Differentiation of Parametric Equations

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