A curve is defined by $x=t+(1/t)$ and $y=t^2+(1/(t^2))$ for $ t≠0.$

Question

A curve is defined by $x=t+(1/t)$ and $ y=t^2+(1/(t^2))$ for $t≠0. $

a) Find the cartesian equation of the curve. b) By considering discriminant, or otherwise, find values of k for which x=k has solutions, where k is constant. c) Sketch the curve, showing any domain restrictions implied by the above parts.

I have figured out part a i.e y=x^2-2 but I don't know how to do part b which is also required to do part c. Any help would be appreciated :) Thank you!

Answer 1

$x^2=y-2$ is the parabola with vertex (0,2). Domain restriction is $y\geq 2.$ Can you draw it?

Answer 2

OP is right yje equation of the curve is a parabola $x^2=y-2$, this is vertical upward parabola with vertex at $(0,2)$. Comparing it with $X^2=4AY$, its ;ength of latus rectum is 1$(=4A)$ unit. Focus is given by $Y=A \implies y=2+1/4, x=0, F=(0,9/4)$. yocan noe plot it .

Answer 3

Setting $x$ equal to $k$ you get $k = t + \dfrac{1}{t}$, which if you multiply through by $t$ gives

$$ k t = t^2 +1$$

which is a quadratic equation in $t$. Do you know about the quadratic formula, and how it involves the discriminant, and what the sign of the discriminant tells you about whether or not the quadratic equation has (real) roots? If you put that equation in standard quadratic form you should be able write a formula for the discriminant that involves $k$, which will tell you which values of $k$ will let you find a corresponding value of $t$.

Answer 4

Well can you please be clear whether '$\ x=k$ has solutions' or '$\ x=k$ is solution'? Because if '$\ x=k$ has solutions' is what you actually meant then it doesn't make any sense, as $\ x=k$ is a straight line perpendicular to the $\ x$-axis and passing through $\ (k,0)$. However I proceed assuming that you want a solution of the given curve.
So as you correctly find out the cartesian form of the given parametric equation which is $\ y=x^2-2$, we move to find out the solutions of this curve, i.e the points where this curve meets the $\ x$-axis.
Now $\ y=0\implies x^2-2=0\implies x=\pm \sqrt{2}$. So the required values are $\ k=\pm\sqrt{2}$
You can see the graph of this function which is a parabola here. The parabola has its vertex at $\ (0,-2)$, latus rectum is $\ 1$ and focus is at $\ (0,-\frac{7}{4})$.