JEE Mathematics 2023 Question with Solution

Given: The curve C satisfies $\frac{dy}{dx} + \frac{x+a}{y-2}=0$ with $y(1)=0$, and the area enclosed is $4\pi$.

Find: The length of the segment RS formed by the intersections of the normals at the y-axis points P and Q with the x-axis.

From the solution,

which gives

Hence,

Integrating,

Using $y(1)=0$, the solution gives

Therefore the curve is written as

The enclosed area is given as $4\pi$. Treating the curve as a circle, the solution uses

so

Then it concludes

Hence,

Substituting, the points of intersection with the y-axis are obtained as

The normals at these points are taken as

and

Setting $y=0$ to find the x-axis intercepts,

Therefore,

the solution explicitly concludes this as

which matches option A on the solution, although it disagrees with the listed option numbering in the answer key. Using the solution as authority, the correct option is A.

Given: $\frac{dy}{dx} + \frac{x + a}{y - 2} = 0$ and $y(1)=0$.

Find: The distance between the x-axis intercepts of the normals at the two y-axis intersection points.

Rearrange the differential equation as used in the extracted working:

Integrating gives an equation of a circle in the form

Comparing with the general circle equation,

we have

So the radius satisfies

Since the area is $4\pi$,

Hence,

so

that is,

Therefore,

Now the circle becomes

Completing squares,

Its center is $(1,2)$ and radius is $2$.

On the y-axis, put $x=0$:

so

and therefore

The normal to a circle at any point passes through its center. Thus the normals at P and Q are the lines through $(1,2)$ and each of these points. Their slopes are $\pm\sqrt{3}$, so the equations are

and

Setting $y=0$ gives x-axis intercepts whose separation is

Therefore, the correct option according to the extracted the solution is A.