Animated Solution for Mathematics - Conic Sections: If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse $x^2+4y^2=36$ is $r$, then $12r^2$ is equal to

The ellipse is defined by the equation $x^2+4y^2=36$. To standardize this, we divide the entire equation by $36$:

Comparing this to the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we identify the parameters $a^2=36$ (so $a=6$) and $b^2=9$ (so $b=3$). We are dealing with a horizontal ellipse stretched along the $x$-axis.

The Normal

To find the largest circle centered at $C(2,0)$ that fits inside the ellipse, we must identify the point of contact $P$ where the circle is tangent to the ellipse. At this point, the normal to the ellipse must pass through the center of the circle, $C(2,0)$.

The normal to the ellipse at any parametric point $(a\cos \theta ,b\sin \theta )$ is given by:

Substituting our values $a=6$ and $b=3$, the equation becomes:

Since the normal must pass through the center $C(2,0)$, we substitute $x=2$ and $y=0$ into the normal equation:

The term involving $\csc \theta$ vanishes, leaving us with $12\sec \theta =27$. Solving for $\sec \theta$, we find:

Now, we determine the coordinates of the contact point $P(x_P,y_P)$:

The radius $r$ of the circle is the distance between $C(2,0)$ and $P(\frac{8}{3},\frac{\sqrt{65}}{3})$. Using the distance formula $r^2=(x_P-2{)}^2+(y_P-0{)}^2$:

The problem asks for the value of $12r^2$:

The final result is 92.