Coordinate Geometry – Parabola – JEE Main 29 Jan 2025 Shift 1

Solution:

Let the focus be $S(4,3)$.

For Parabola I, the directrix is the x-axis ($y=0$). The equation is:

$(x-4)^{2} + (y-3)^{2} = y^{2}$ …..(1)

For Parabola II, the directrix is the y-axis ($x=0$). The equation is:

$(x-4)^{2} + (y-3)^{2} = x^{2}$ …..(2)

The intersection points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ satisfy both equations.

From (1) and (2), we equate the left-hand sides:

$y^{2} = x^{2} \Rightarrow y = x$ (Since the focus is in the first quadrant and directrices are axes, the intersection lies where coordinates have same sign).

Substitute $y=x$ into equation (1):

$(x-4)^{2} + (x-3)^{2} = x^{2}$

$x^{2} – 8x + 16 + x^{2} – 6x + 9 = x^{2}$

$x^{2} – 14x + 25 = 0$

Let roots be $x_{1}$ and $x_{2}$.

$x_{1} + x_{2} = 14$

$x_{1}x_{2} = 25$

The points are $A(x_{1}, x_{1})$ and $B(x_{2}, x_{2})$ since $y=x$.

The distance squared $(AB)^{2}$ is:

$(AB)^{2} = (x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2}$

$(AB)^{2} = (x_{1}-x_{2})^{2} + (x_{1}-x_{2})^{2} = 2(x_{1}-x_{2})^{2}$

Using the identity $(x_{1}-x_{2})^{2} = (x_{1}+x_{2})^{2} – 4x_{1}x_{2}$:

$(AB)^{2} = 2[(14)^{2} – 4(25)]$

$(AB)^{2} = 2[196 – 100]$

$(AB)^{2} = 2[96] = 192$

Ans. (1)