AOD – Maxima and Minima – JEE Main 2025 Shift 2

Solution:

Let $P(x, y)$ be a point on the parabola $y^2 = 4x$. We can parameterize $P$ as $(t^2, 2t)$.
Let $A = (a, 0)$. The distance squared between $P$ and $A$ is function $f(t)$:
$$D^2 = f(t) = (t^2 – a)^2 + (2t – 0)^2$$
For the shortest distance (minima), we differentiate with respect to $t$ and set to 0 ($f'(t) = 0$):
$$f'(t) = 2(t^2 – a)(2t) + 2(2t)(2) = 0$$
$$4t(t^2 – a + 2) = 0$$
Since $P$ is not the vertex (given shortest distance is 4, not $a$), $t \ne 0$. Thus:
$$t^2 – a + 2 = 0 \Rightarrow a = t^2 + 2$$

We are given the minimum distance is 4. So at this $t$:
$$(t^2 – a)^2 + (2t)^2 = 4^2$$
Substitute $(t^2 – a) = -2$:
$$(-2)^2 + 4t^2 = 16$$
$$4 + 4t^2 = 16$$
$$4t^2 = 12 \Rightarrow t^2 = 3$$

Now find $a$:
$$a = t^2 + 2 = 3 + 2 = 5$$
The point $(a, 0)$ is $(5, 0)$.
The focus of $y^2 = 4x$ is $(1, 0)$.

We need the circle passing through $(5, 0)$ and $(1, 0)$ with the centre on the x-axis.
Since the centre is on the axis joining these two points, $(1, 0)$ and $(5, 0)$ are the endpoints of the diameter.
Equation of the circle:
$$(x – 1)(x – 5) + (y – 0)(y – 0) = 0$$
$$x^2 – 6x + 5 + y^2 = 0$$
$$x^2 + y^2 – 6x + 5 = 0$$

Ans. (1)