We have to find the minimum and maximum distance of a point $(9/5 ,12/5)$ from the ellipse $4(3x+4y)^2 +9(4x-3y)^2 =900$ .
I got to know the center of ellipse would be $(0,0)$ .
after that I am not getting any start.
I think there would be some trick to solve it .
Introduce the following variables (Leibovici style$^*$)
$$u=2(3x+4y)\,\,\text{ and } \,\, v=3(4x-3y).$$
Then we get a circle of radius $30$ in the $u,v$ coordinate system:$$u^2+v^2=900.$$
In this coordinate system the point from which we measure the distance of the points on the circle is
$$(30, 0).$$
So, the closest point on the circle to this point is at $(30,0)$ and the farthest point on the circle is at $(-30,0)$.
The minimum distance is $0$ and the corresponding point on the ellipse is $\left(\frac95,\frac{12}5\right)$.
For the other (farthest) point, use the inverse of the transformation we used to get the coordinates of the point $(-30,0)$ points in the $(x,y)$ system. Or solve the following system of equations
$$-30=6x+8y\ \ \\ \ \ \ \ \ 0=12x-9y.$$
Then calculate the distance of the two points that we have just found.
EDIT
The solution is $\left(-\frac95,-\frac{12}5\right)$. The closest point is $\left(\frac95,\frac{12}5\right)$. The distance is
$$\sqrt{\left(\frac{18}5\right)^2+\left(\frac{24}5\right)^2}=6.$$
$^*$Claude Leibovici suggested in a later deleted comment the transformation
$$u=3x+4y\,\,\text{ and } \,\, v=4x-3y.$$
1) That formula is kinda hard to work with. You should start by rewriting it to collect all the $x$ terms together and then collect all the $y$ terms together.
2) the simplest way to solve problems like this tends to be basic calculus! Write down the formula for the distance from a point on the ellipse to your point, and differentiate!
Hint...try rotating the ellipse by using $u=\frac{3x+4y}{5}$ and $v=\frac{4x-3y}{5}$ and transforming the given fixed point
The point $(\frac 95,\frac{12}5)$ is on the ellipse $$4(3x+4y)^2 +9(4x-3y)^2 =900\tag1$$
So, the minimum distance is $\color{red}0$.
We want to maximize $$D=\bigg(X-\frac 95\bigg)^2+\bigg(Y-\frac{12}5\bigg)^2$$ under the condition that $$4(3X+4Y)^2 +9(4X-3Y)^2 =900\tag2$$
Here, $(2)$ can be written as $$\frac{(3X+4Y)^2}{15^2}+\frac{(4X-3Y)^2}{10^2}=1$$
So, we can set $$3X+4Y=15\cos t,\qquad 4X-3Y=10\sin t$$ Solving this for $X,Y$ gives $$X=\frac 95\cos t+\frac 85\sin t,\ \ Y=\frac{12}{5}\cos t-\frac 65\sin t$$
Using these, we have $$\begin{align}D&=\bigg(\frac 95\cos t+\frac 85\sin t-\frac 95\bigg)^2+\bigg(\frac{12}{5}\cos t-\frac 65\sin t-\frac{12}5\bigg)^2 \\\\&=4\sin^2t+9\cos^2t-18\cos t+9 \\\\&=5\cos^2t-18\cos t+13 \\\\&=5\bigg(\cos t-\frac 95\bigg)^2-\frac{16}{5}\end{align}$$ which is an upward parabola of $\cos t$ where $-1\le \cos t\le 1$.
So, we see that $D$ is maximized when $\cos t=-1$ for which we have $D=36$.
Therefore, the maximum distance is $\sqrt{36}=\color{red}6$ which is attained when $(X,Y)=(-\frac 95,-\frac{12}{5})$.
The given conic equation is
$ 4(3x + 4 y)^2 + 9(4x - 3 y)^2 = 900 $
Let's define the rotation matrix
$ R = \begin{bmatrix} \dfrac{3}{5} && \dfrac{-4}{5} \\ \dfrac{4}{5} && \dfrac{3}{5} \end{bmatrix} $
And let $r = [x, y]^T $ , and define $z = [z_1, z_2]^T = R^T r $ (so that $r = R z $), then
$ z = \dfrac{1}{5} \begin{bmatrix} 3 x + 4 y \\ - 4 x + 3 y \end{bmatrix} $
Therefore,
$ 4 z_1^2 + 9 z_2^2 = \dfrac{900}{25} = 36 $
In the $Oz_1 z_2 $ frame, this is an ellipse with a horizontal ($z_1$) semi-major axis length of $3$ and a vertical ($z_2$) semi-minor axis length of $2$.
The point $ (\dfrac{9}{5}, \dfrac{12}{5} ) $ corresponds to
$ (z_1 , z_2 ) = R^T r = \dfrac{1}{25} ( 75, 0 ) = (3, 0)$
This point is the "right" vertex of the ellipse. So it is ON the ellipse, and hence the minimum distance from this point to the ellipse is zero.
The maximum distance between this point and (another) point on the ellipse is obtained with the opposite vertex (which is the "left" vertex) $(-3, 0)$.
The maximum distance is the length of the major axis, which is $2 (3) = 6 $.
And the coordinates $(x,y)$ of this points is given by
$ \begin{bmatrix}x \\y \end{bmatrix} = R \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \dfrac{1}{5} \begin{bmatrix} 3 && - 4 \\ 4 && 3 \end{bmatrix} \begin{bmatrix} -3 \\ 0 \end{bmatrix} = \begin{bmatrix} - \dfrac{ 9 }{5} \\ -\dfrac{12}{5} \end{bmatrix} $
That is, the maximizing point is $( - \dfrac{9}{5}, -\dfrac{12}{5} ) $
