Added: In a comment OP states that "The major axis is on the y-axis and the minor axis is on the x-axis."
The equation of an ellipse whose major and minor axis are respectively on
the $y$ and $x$-axis is
$$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad (\ast )$$
where $a$ is the semimajor axe and $b$ is the semiminor axe. You are given $%
2a$ and you need to find $2b$. Let the coordinates of the given point be $%
(x_{1},y_{1})$. Since it is on the ellipse, its coordinates must satisfy $%
(\ast )$
$$\frac{x_{1}^{2}}{b^{2}}+\frac{y_{1}^{2}}{a^{2}}=1.\qquad (\ast \ast )$$
Clearing denominators and then dividing by $y_{1}^{2}-a^{2}$ we get
$$a^{2}x_{1}^{2}+b^{2}y_{1}^{2}=a^{2}b^{2}\Leftrightarrow \left(
y_{1}^{2}-a^{2}\right) b^{2}=-a^{2}x_{1}^{2}\Leftrightarrow b^{2}=-\frac{%
a^{2}x_{1}^{2}}{y_{1}^{2}-a^{2}}=\frac{a^{2}x_{1}^{2}}{a^{2}-y_{1}^{2}}.$$
Since $a^{2}-y_{1}^{2}\geq 0$ and $b>0$, we obtain
$$b=\frac{a|x_{1}|}{\sqrt{a^{2}-y_{1}^{2}}}.\qquad (\ast \ast \ast )$$
The length of the minor axe is $2b$.
Let $D(x_d,y_d)$ and $k$ the slope of the tangent. The ellipse has the form,
$$\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} = 1$$
Its derivative is $y'=-\frac{b^2x}{a^2(y-b)}$ and matches $k$ at D,
$$k = -\frac{b^2x_d}{a^2(y_d-b)} \tag{1}$$
The point $(x_d,y_d)$ satisfies,
$$\frac{x_d^2}{a^2} + \frac{(y_d-b)^2}{b^2} = 1\tag{2}$$
Combine (1) an (2) to get the following equation for $b$,
$$(y_d-b)^2-kx_d(y_d-b)=b^2$$
which yields the unique solution for the minor axis $b$,
$$b= \frac{y_d(y_d-kx_d)}{2y_d-kx_d}$$
Then, plug the solution $b$ into (1) to get the major axis $a$.
Best Answer
Let's suppose it was semi-major axis $4k$ and semi-minor axis $k$ to avoid confusion.
Aligned with the axes it would be
$$\frac{x^2}{(4k)^2}+\frac{y^2}{k^2}=1$$
but you want this rotated, so replace $x$ by $\frac{x+y}{\sqrt 2}$ and $y$ by $\frac{y-x}{\sqrt 2}$ to get
$$\frac{(x+y)^2}{32k^2}+\frac{(y-x)^2}{2k^2}=1$$ which you can also write as
$$17\,{y}^{2}-30\,x\,y+17\,{x}^{2} - 32\,{k}^{2} = 0.$$