Find the line tangent to $9x^2+16y^2=52$ that is parallel to $9x-8y=5$ using the first order Taylor Series 
Find the line tangent to $9x^2+16y^2=52$ that is parallel to $9x-8y=5$

By setting $z(x,y) =0$, I used the tangent plane formula or first order TS
$$
h(x,y)=z(a,b)+z_x(a,b)(x-a)+z_y(a,b)(y-b)
$$
$$
h(x,y)=9a^2+16b^2-52+18a(x-a)+32b(y-b)
$$
$$
h(x,y)=-9a^2-16b^2-52+18ax+32by
$$
To make the plane parallel to $9x-8y=5$ at $z=0$, I set
$$
a=1/2\\
b=-1/4
$$
to take the form of the parallel line which then gives me a plane
$$
h(x,y)= 9x-8y -55.25
$$
that is not tangent to the $9x^2+16y^2=52$ but parallel to $9x-8y=5$ at $h(x,y)=0$
What have gone wrong with my solution?
Edit:
If I got $a,b$ correct, the graph should look similar to this and I will simply take $h(x,y)=0$ to get the line I need

 A: You should not be using the tangent plane formula. That is, as the name implies, used for finding tangent planes (so, a 2-dimensional object in 3-dimensional space). What you're looking for is a tangent line which is much different, so when you define your function $z(x,y)$, it actually doesn't at all mean what you want it to and does not define an ellipse. Here's roughly what it looks like:

(https://www.desmos.com/calculator/xyrgbanmmi)
You really just want to represent the two halves of the ellipse in two separate equations and solve it using single variable calculus. $9x^2+16y^2=52$ can be rewritten as $y^2=\frac{-9x^2+52}{16}$ which is $y=\pm\frac{\sqrt{-9x^2+52}}{4}$.
Now, if you write those two parts out separately, you can get the derivative and just find points where the slope of the tangent is equal to your given line and find the equation. You should be getting two tangent lines, since the tangent line at any point on an ellipse has the same slope as the tangent line to the point directly opposite it.
There's probably a more elegant way to represent this and solve for a derivative / obtain the points, but in any case, you definitely should not be using a formula designed for three dimensions. $z(x,y)$ is supposed to be a function that takes the $x$- and $y$-coordinate and return the $z$-coordinate at that point, not a relation that determines if a point in 2D is on a figure or not.
