Let's practise the basics, Part 2

Brian has written a fine solution. For the sake of variety, let me propose a solution in terms of quadratic forms.

The symmetric matrix of our quadratic form $36x^2+24xy+29y^2$ is $A=\begin{bmatrix}36&12\\12&29\end{bmatrix}$

with characteristic polynomial $p(\lambda)=\lambda^2-65\lambda+900=(\lambda-20)(\lambda-45)$ so that the standard form of the ellipse is $20u^2+45v^2=\frac{180}{\pi}$ or $\frac{\pi u^2}{9}+\frac{\pi v^2}{4}=1$ . The semiaxes are $a=\frac{3}{\sqrt{\pi}}$ and $b=\frac{2}{\sqrt{\pi}}$ so that the area is $\pi ab=\boxed{6}$

While it is easy to find the eigenvalues and the semi-axes in this case, it is not logically necessary; the area will be $\frac{180}{\sqrt{\det A}}=6$

The given equation is that of a rotated ellipse. To eliminate the $xy$ -term and create an equation of an ellipse in standard form, (so that we can employ the ellipse area formula $\pi ab$ ,) we will need to execute a rotation of axes as per the method outlined in this link .

As described in the link, given an equation $Ax^{2} + Bxy + Cy^{2} + Dx + Ey = F,$ we can, through a rotation of axes to an $x',y'$ grid, create a transformed equation $A'(x')^{2} + C'(y')^{2} + D'x' + E'y' = F'$ where

$A' = A\cos^{2}(\alpha) + B\cos(\alpha)\sin(\alpha) + C\sin^{2}(\alpha),$

$C' = A\sin^{2}(\alpha) - B\sin(\alpha)\cos(\alpha) + C\cos^{2}(\alpha),$

$D' = D\cos(\alpha) + E\sin(\alpha), E' = -D\sin(\alpha) + E\cos(\alpha), F' = F,$

where $\cot(2\alpha) = \dfrac{A - C}{B}$ such that $0 \lt \alpha \lt 90^{\circ}.$

Now with $A = 36, B = 24, C = 29, D = E = 0$ and $F = \dfrac{180}{\pi},$ we have that

$\cot(2\alpha) = \dfrac{36 - 29}{24} = \dfrac{7}{24} \Longrightarrow \cos(2\alpha) = \dfrac{7}{25}$

$\Longrightarrow \cos(\alpha) = \sqrt{\dfrac{1 + \cos(2\alpha)}{2}} = \sqrt{\dfrac{1 + \frac{7}{25}}{2}} = \dfrac{4}{5}$ and $\sin(\alpha) = \dfrac{3}{5}.$

Plugging in the appropriate values, we find that $A' = 45, C' = 20, D' = E' = 0$ and $F' = \dfrac{180}{\pi}.$ Our transformed equation then becomes

$45(x')^{2} + 20(y')^{2} = \dfrac{180}{\pi} \Longrightarrow \dfrac{(x')^{2}}{\dfrac{4}{\pi}} + \dfrac{(y')^{2}}{\dfrac{9}{\pi}} = 1 \Longrightarrow \dfrac{(x')^{2}}{a^{2}} + \dfrac{(y')^{2}}{b^{2}} = 1$

where $a = \dfrac{2}{\sqrt{\pi}}$ and $b = \dfrac{3}{\sqrt{\pi}}.$ The area of this ellipse in then $\pi ab = \pi * \dfrac{2}{\sqrt{\pi}} * \dfrac{3}{\sqrt{\pi}} = \boxed{6}.$

Brian Sir's approach is nice . i did not thought of rotating the coordinate axes .

we can figure out the given curve is an ellipse . (using the conditions for an equation to represent an ellipse)

Partially differentiating the equation once wrt x and then wrt y and then solving them simultaneously we

will obtain the coordinates of the centre of ellipse which turns out to be (0,0) .

now we assume a variable chord of the ellipse y = mx passing through the centre .

for the chord to be major or minor axis the distance of point of intersection of the chord with the ellipse

from the centre should be maximum or minimum respectively . (as any chord passing through the centre of

ellipse is bisected at the centre ) .

using parametric coordinates of a straight line we assume a point (rcosx , rsinx) on the line which lies at a

distance of r from the centre where x is the angle made by the line with positive direction x axis .

now substituting this point in the equation of ellipse

we will obtain an expression of r in terms of the angle x .

we can easily maximise or minimise that expression using trigonometry to get the value of semi major or

minor axis .

I am introducing an ordinary method here. But don't think about finding an area of an ellipse. Rearrange:

$29 y^2 + 24 x y + 36 x^2 - \frac{180}{\pi}= 0$

$\Rightarrow y^2 + \frac{24}{29} x y + \frac{36}{29} x^2 - \frac{180}{29 \pi}= 0$

$\Rightarrow y^2 + 2 (\frac{12}{29}) x y + \frac{36}{29} x^2 - \frac{180}{29 \pi}= 0$

Obtain top curve and bottom curve from rectified quadratic forms and simplified :

$y_1 = \frac{-12}{29} x + \sqrt{\frac{180}{29 \pi} - \frac{900}{841} x^2}$ and

$y_2 = \frac{-12}{29} x - \sqrt{\frac{180}{29 \pi} - \frac{900}{841} x^2}$

$\frac{180}{29 \pi} - \frac{900}{841} x^2$ $\geq$ 0

$\Rightarrow -\sqrt \frac{29}{5 \pi} \leq x \leq \sqrt \frac{29}{5 \pi}$

Plot the graphs to observe as follows:

To be more familiar with curve above x-axis (or curve below x-axis), we add 2 to both $y_1$ and $y_2$ :

$y_{top} = 2 - \frac{12}{29} x + \sqrt{\frac{180}{29 \pi} - \frac{900}{841} x^2}$ and

$y_{bottom} = 2 - \frac{12}{29} x - \sqrt{\frac{180}{29 \pi} - \frac{900}{841} x^2}$

Let $x$ = $\sqrt{\frac{841}{900}\times \frac{180}{29 \pi}} \sin \theta$ = $\sqrt{\frac{29}{5 \pi}} \sin \theta$

$d x$ = $\sqrt{\frac{29}{5 \pi}} \cos \theta$ $d \theta$ and range becomes $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.$

Area enclosed = $\Delta = \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (y_{top} - y_{bottom}) d \theta$

$\Rightarrow \Delta = \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 2 \sqrt {\frac{180}{29 \pi}}\sqrt {\frac{29}{5 \pi}} \cos ^2 \theta$ $d \theta$ {Only a very simple integral wanted to be determined.}

$\Rightarrow \Delta = \displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{12}{\pi} \cos ^2 \theta$ $d \theta$

$\Rightarrow \Delta = \displaystyle \frac{24}{\pi}\displaystyle \int_0^{\frac{\pi}{2}} \cos ^2 \theta$ $d \theta$

$\Rightarrow \Delta = \displaystyle \frac{24}{\pi} \displaystyle \frac{\pi}{4} = 6$

$Note:$ Transformation onto the ellipse into rectangular orientation is $NOT$ practical as major and minor of it are in terms of $\pi$ and irrational, while the angle of turning required is also not convenient to determine for irrational numbers. Where this was also done by others, deeper geometry concerning rotation ought to involve. Where as here, we don't employ $\pi$ a b.

Answer: $\boxed{6}$