- #1

W3bbo

- 31

- 0

## Homework Statement

Reduce these parametric functions to a single cartesian equation:

[tex]

$\displaylines{

x = at^2 \cr

y = 2at \cr} $

$\displaylines{

x = 3{\mathop{\rm Sec}\nolimits} \left( \alpha \right) \cr

y = 5{\mathop{\rm Tan}\nolimits} \left( \alpha \right) \cr} $

$\displaylines{

x = t^2 + 1 \cr

y = t^2 + t \cr} $

[/tex]

## Homework Equations

N/A

## The Attempt at a Solution

For Q1, I think I got the hang of it... in that the plots of my cartesian and the paremetric are identical:

[tex]

$\displaylines{

x = at^2 \cr

y = 2at \cr

\cr

t^2 = {x \over a} \cr

t = \sqrt {{x \over a}} \cr

\cr

y = 2a\sqrt {{x \over a}} \cr} $

[/tex]

For Q2 I thought I had it solved with a trigonometric identity, but the plots look different:

[tex]

$\displaylines{

x = 3{\mathop{\rm Sec}\nolimits} \left( \alpha \right) \cr

y = 5{\mathop{\rm Tan}\nolimits} \left( \alpha \right) \cr

\cr

{\mathop{\rm Sec}\nolimits} ^2 \left( \alpha \right) = 1 + {\mathop{\rm Tan}\nolimits} ^2 \left( \alpha \right) \cr

\left( {{\textstyle{1 \over 3}}x} \right)^2 = 1 + \left( {{\textstyle{1 \over 5}}y} \right)^2 \cr

{\textstyle{1 \over 9}}x^2 = 1 + {\textstyle{1 \over {25}}}y^2 \cr

{\textstyle{1 \over {25}}}y^2 = {\textstyle{1 \over 9}}x^2 - 1 \cr

y^2 = {\textstyle{{25} \over 9}}x^2 - 1 \cr

y = \pm \sqrt {{\textstyle{{25} \over 9}}x^2 - 1} \cr} $

[/tex]

Finally, Q3 seems deceptivly simple, but again, the plot doesn't match the original:

[tex]

$\displaylines{

x = t^2 + 1 \cr

y = t^2 + t \cr

\cr

t^2 = x - 1 \cr

y = x - 1 + \sqrt {x - 1} \cr} $

[/tex]

I'm not looking for answers, just to find out where I've gone wrong.

Thanks