While solving problems in my preparation for exam I faced parametric equation related to calculus.
I can deal with finding tangent line but I am in no way being able to relate x to y by solving for t in both equations. It seems that there is a trick that I am not being able to see. Can some one help me with this by just giving direction.
On
The formula for slope with parametric equations is
$${dy\over dx}= {{dy\over dt}\over{dx\over dt}}$$
If you can get the derivatives ${dy\over dt}$and${dx\over dt}$, you will have ${dy\over dx}$ in terms of $t$ and you would plug in $t=2$ to get the slope of the tangent line. The equation would then be found using the point-slope formula.
While I do suggest you use
$\frac {dy}{dx} = \frac{\frac {dy}{dt}}{\frac {dx}{dt}}$
This is just an application of the chain rule.
It is possible to put this into a more familiar form:
$9x^2 + y^2 =$$ \frac {9t^2}{(1+t^2)^2} + \frac {4-4t^2+t^4}{(1+t^2)^2}\\ \frac {4+5t^2 + t^4}{(1+t^2)^2} = \frac {(t^2 + 1)(4t^2 + 1)}{(1+t^2)^2}\\ \frac {(4t^2 + 1)}{(1+t^2)}\\ \frac {-(1-2t^2) + 2 (1+t^2)}{(1+t^2)} = -y + 2$
$9x^2 + y^2 = -y + 2$
Tells us that our parameterziation gives us an ellipse.