JEE Main 2023 — Ellipse Question with Solution

The given ellipse has the equation :

$$x^2+4y^2=36$$

We can rewrite this as :

$$\frac{x^2}{6^2}+\frac{y^2}{(6/2{)}^2}=1$$

This shows that it is an ellipse centered at (0,0) with semi-major axis a = 6 along the x-axis and semi-minor axis b = 3 along the y-axis.

The equation of a circle with center (2,0) and radius r is :

$$(x-2{)}^2+y^2=r^2$$

Substituting y^2 from the ellipse equation into the circle equation gives us :

$$x^2-4x+4+\frac{36-x^2}{4}=r^2$$

Solving this equation leads to :

$$3x^2-16x+52-4r^2=0$$

For the roots of this quadratic equation to be real (which they must be, since they represent real intersection points), the discriminant (D) must be greater than or equal to zero :

$$D=b^2-4ac=(-16{)}^2-4\times 3\times (52-4r^2)=256-12\times 52+48r^2$$

Setting D = 0 gives the minimum value for r (the radius of the inscribed circle) :

$$256-624+48r^2=0$$

$$48r^2=368$$

$$r^2=\frac{368}{48}=\frac{23}{3}$$

And we're asked for the value of 12r², so :

$$12r^2=12\times \frac{23}{3}=92$$

So, the correct answer is Option B : 92.