Given the equation of ellipse $\frac{x^2}{9} +y^2 =1 $.
Find the tangent at P $(\frac{3}{\sqrt2},\frac{1}{\sqrt2})$.$\\$
I know how to find tangent plane but I need an answer in parametric form like the options given ,which are q(w) = $[\frac{1}{\sqrt2}(1-w),\frac{3}{\sqrt2}(1+w)]$. Any hints?
On
The standard way is:
For poin P
$$ x= 3 \cos t, y=\sin t $$
For the given point parameter $t=\pi/4$ slope
$$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}= \dfrac{-1}{3 \tan t}$$
Point tangent slope form
$$\dfrac{y-1/\sqrt 2}{x-3/\sqrt 2}=\dfrac13\dfrac{\sin t-1/\sqrt 2}{\cos t-1/\sqrt 2}= \dfrac{-1}{3 \tan t}$$
You have not defined what w is. Incorporate the same into the above tangent equation.
Guide:
After you find the tangent in the form of $y=mx+c$ where $m$ and $c$ are known.
If you just want any particular parametric form, you can let $x=t$, then $y=mt+c$ and you can write it as $[t, mt+c]$.
If you are given options $[x,y]$, try to substitute the $x$ expression into $mx+c$ and verify if you get the corresponding $y$.