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the parametric equations of a curve are $$ \left\{ \begin{array}{ll} x =&\alpha (t- 1/t) \\ y=&\alpha(t+ 1/t) \end{array} \right. $$ where $\alpha$ is a constant. find the gradient of the tangent to the curve at the point where $t=2$. hence, obtain the eqn of the normal at this point.

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  • $\begingroup$ What have you tried so far? How would you calculate the gradient? $\endgroup$ – Matti P. Feb 11 at 11:14
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First make partial derivative $\frac{\partial x}{\partial t}= \alpha (1+\frac{1}{t^2})$ and $\frac{\partial y}{\partial t}= \alpha (1-\frac{1}{t^2})$. Now $t=2$ so the gradient is substitude $t=2$ in the partial derivatives, so the gradient is $\alpha(\frac{5}{4},\frac{3}{4})$. Now the slope of the tangent line in $t=2$ is $m = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}$. You want the normal line in $t=2$ so the slope of the normal given the slope of the tangent is $n=-\frac{1}{m}$, so $n=-\frac{5}{3}$. Now using the formula for a line $y-y_0 = m(x-x_0)$, where in your case $(x_0,y_0)=\alpha(\frac{3}{2},\frac{5}{2})$ so the normal line in $t=2$ is $y-\alpha\frac{5}{2} = -\frac{5}{3}(x-\alpha\frac{3}{2})$

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