How to give the sketch of a set

Let $x=rcos t, y=r sin t$, so $$r^2(t)=frac{72}{13-10 sin t cos t}=frac{144}{26-10 sin 2t}.$$ So $$r^2(t)_{min}=r^2(t=3pi/4)=frac{144}{36},~r^2(t)_{max}=r^2(t=pi/4)=frac{144}{16}$$ $r_{min}=2, r_{max}=3$, these are the semi-minor and semi-major axes of an ellipse which are inclined at an angle $3pi/4$ and $pi/4$ with x-axis.

Analyze the given relation

• It is symmetric about $x=y$ as interchanging $x$ and $y$ yield the same equation
• It is symmetric about $x+y=0$ replacing $y$ by $-x$ and $x$ by $-y$ yield the same equation.
• It is a 2-degree equation in $x$ and $y$. So it must be a conic.
• It cannot be a circle beacuse the coefficient of $xy$ is non-zero
• It cannot be a parabola because a parabola has only one axis of symmetry whereas the curve already has two.
• The coefficients of $x^2$ and $y^2$ are of the same sign. So it cannot be a hyperbola as well.
• So the equation must represent an ellipse, with the axes being $x=y$ and $x+y=0$. So the centre is at $(0,0)$

So you need to find the maximum and minimum distances of a point on an ellipse from its centre. Evidently the points will be along the axes. So solve the given equation with $x=y$ and $x+y=0$ to get the points. One of them will represent the maximum distance and the other one will represent the minimum distance. The rest of the work is left to the reader.

Finally, this is how the ellipse looks like

The form of the equation should indicate to you that this is an ellipse with axes $45^circ$ rotated from the standard axes. As such, the LHS can be rewritten with $(x+y)^2$ and $(x-y)^2$: $$9(x-y)^2+4(x+y)^2=72$$ $$frac{(x-y)^2}8+frac{(x+y)^2}{18}=1$$ Therefore the nearest and farthest points from the origin correspond to vertices of the ellipse and have coordinates of the form $(x,pm x)$; you should find that the semi-major and semi-minor axes have lengths $3$ and $2$ respectively, with the semi-major axis parallel to $x=y$. From there you should be able to make a sketch.