91Ó°ÊÓ

We start by diagonalizing the quadratic form, which allows us to rewrite the equation in the standard form of an ellipse. The eigenvalues of the matrix associated with the quadratic form can be found by solving the following equation for $$\lambda$$: $$\begin{vmatrix} a-\lambda & b \\ b & c-\lambda \\ \end{vmatrix}=0$$ which gives us $$(a-\lambda)(c - \lambda) - b^2 = \lambda^2 - (a+c)\lambda + (ac - b^2) = 0$$ The discriminant is $$(a+c)^2 - 4(ac - b^2) = a^2 - 2ac + c^2 + 4b^2 > 0$$ since $$ac - b^2 > 0$$ and $$a,c > 0.$$ Therefore, we have two distinct eigenvalues $$\lambda_1, \lambda_2 > 0.$$ We can rotate the coordinate axes by some angle $$\theta$$ to diagonalize the quadratic form to find the corresponding eigenvectors.

When we diagonalize the quadratic form using the eigenvalues $$\lambda_1$$ and $$\lambda_2$$, we obtain the following diagonalized equation for the ellipse: $$\lambda_1 x'^2 + \lambda_2 y'^2 = 1$$ Now we can find the principal axes. The lengths of the semi-major and semi-minor axes are given by: $$a' = \frac{1}{\sqrt{\lambda_1}}$$ $$b' = \frac{1}{\sqrt{\lambda_2}}$$

Now that we have the lengths of the semi-major and semi-minor axes, we can calculate the area of the elliptical region using the standard formula for the area of an ellipse: $$A = \pi a'b'$$ Substitute the values of $$a'$$ and $$b'$$ in this equation: $$A = \pi \left(\frac{1}{\sqrt{\lambda_1}}\right)\left(\frac{1}{\sqrt{\lambda_2}}\right) = \frac{\pi}{\sqrt{\lambda_1 \lambda_2}}$$ Since the product of the eigenvalues is equal to $$\lambda_1 \lambda_2 = ac - b^2$$ (because the determinant of a diagonal matrix is the product of its elements), we can write: $$A = \frac{\pi}{\sqrt{ac-b^2}}$$ Thus, we have proved that the area of the elliptical region given by $$ax^2 +2bxy + cy^2 \leq 1$$ is equal to $$\frac{\pi}{\sqrt{ac-b^2}}.$$