$\text{The equation of the ellipse whose axes are coincident with the coordinates axes and which touches the straight lines}$

$3x-2y-20=0\text{ and }x+6y-20=0\text{ is }$

$\text{Let the equation of the ellipse be}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\text{We know that the general equation of the tangent to the ellipse is}$

$y=mx\pm \sqrt{a^2{m}^2+b^2}----\left(i\right)$

$\begin{array}{l}\text{ Since }3x-2y-20=0\text{ or }y=\frac{3}{2}x-10\\ \text{ is tangent to the ellipse, comparing with (i), }\end{array}$

$\begin{array}{l}m=\frac{3}{2}\text{ and }{a}^2{m}^2+b^2=100\\ \text{ or }{a}^2\times \frac{9}{4}+b^2=100\\ \text{ or }9a^2+4b^2=400-----\left(ii\right)\end{array}$

$\text{ Similarly, since }x+6y-20=0\text{ , i.e., }$

$y=-\frac{1}{6}x+\frac{10}{3}$

$\text{ is tangent to the ellipse, comparing with (i), }$

$\begin{array}{l}m=\frac{1}{6}\text{ and }{a}^2{m}^2+b^2=\frac{100}{9}\\ \text{ or }\frac{a^2}{36}+b^2=\frac{100}{9}\\ \text{ or }{a}^2+36b^2=400-----\left(iii\right)\end{array}$

$\text{ Solving (ii) and (iii), we get }{a}^2=40\text{ and }{b}^2=10\text{ . }$

$\text{ Therefore, the required equation of the ellipse is }$

$\frac{x^2}{40}+\frac{y^2}{10}=1$