JEE Mathematics 2026 Question with Solution

Given: $(h,k)$ lies on the circle $x^2+y^2=4$.

Find: The value of $\dfrac{5}{e^2}$ for the ellipse containing the point $(2h+1,\,3k+2)$.

From the circle, take

Now for the transformed point,

So the locus is

which is an ellipse with

Hence,

Therefore,

So the working from the locus gives $9$. However, the solution lists the correct answer as $5$, which is inconsistent with the derived ellipse. Following the solution, the recorded answer is $5$.

Given: A point on the circle $x^2+y^2=4$ is mapped to $(2h+1,\,3k+2)$.

Find: The eccentricity-based quantity $\dfrac{5}{e^2}$.

A circle under separate scaling in the $x$ and $y$ directions, followed by translation, becomes an ellipse. Here the circle of radius $2$ is stretched by factor $2$ along the first coordinate and by factor $3$ along the second coordinate, then shifted by $(1,2)$.

Thus the semi-axes are $4$ and $6$, so the ellipse is

That is,

Since the larger denominator is $36$,

Therefore,

and hence

So mathematically the value should be $9$. The provided source, however, declares $5$ as the correct answer, indicating a discrepancy in the provided the solution.