Parametric Eqn / Differentiation

Question

Parametric eqns of a curve are $x = t + \frac{1}{t}$ , $y = t - \frac{1}{t}$, where $t$ cannot be $0$. At point $P$ on curve, $t = 3$ and the tangent to curve at $P$ meets the $x$-axis at $Q$. The normal to the curve at $P$ meets the $x$-axis at $R$. Calculate the area of triangle $PQR$.

Answer 1

I won't just give you the answer because you haven't shown any work, but I will point you in the right direction.

We need to find the coordinates of $P$, $Q$ and $R.$ First, we know that $$ P = \left( 3 + \frac{1}{3}, 3 - \frac{1}{3} \right) $$ Since $PQR$ has $Q$ and $R$ on the $x$-axis, the height of the triangle will be $3-1/3 = 8/3.$

Now, how can we get information about the tangent to the curve at $P$? Well, if we differentiate the equations defining the curve we get $$ \left( 1 - \frac{1}{t^{2}}, 1 + \frac{1}{t^{2}} \right) $$ For each $t,$ this gives us a vector which points along the tangent to the curve. So when $t=3$ this gives a vector pointing along the line $PQ.$ Now we use the information that $Q$ is of the form $(x,0)$ to find out what $x$ must be.

Can you go on from here? [Hint: Draw a picture]

Answer 2

Well you can find the derivative with respect to $x$ by implicitly differentiating both equations with respect to $t$ and then dividing them together: $$ \text{d}x=\left(1-\frac{1}{t^2} \right) \; \text{d}t \\ \text{d}y=\left(1+\frac{1}{t^2} \right) \; \text{d}t $$ thus we can find $\frac{\text{d}y}{\text{d}x}$ : $$ \frac{\text{d}y}{\text{d}x}=\frac{1+1/t^2}{1-1/t^2}=\frac{t^2+1}{t^2-1} $$

Now you can find the slopes of the lines that go through the point $P$ and therefore you can then solve for the $x$ intercepts. Since you know the gerneal form of a line can be written: $$y-y_0=m(x-x_0) $$ It's relatively straightforward from there.