# differentiation questions anyone can help me to solve this

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.

• What have you tried so far? How would you calculate the gradient? – Matti P. Feb 11 at 11:14

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 $$y-\alpha\frac{5}{2} = -\frac{5}{3}(x-\alpha\frac{3}{2})$$