JEE Main 2024 — 6 April, Shift 1 — Question 13

Let the variable line of slope $m>0$ pass through the point $(4,-9)$.

The equation of the line is

$$y+9 = m(x-4)$$

$x$-intercept $A$:

Putting $y=0$,

$$9 = m(x-4) \Rightarrow x = 4 + \frac{9}{m}$$ $$OA = 4 + \frac{9}{m}$$

$y$-intercept $B$:

Putting $x=0$,

$$y + 9 = -4m \Rightarrow y = -(9 + 4m)$$ $$OB = 9 + 4m$$

Sum of distances from the origin:

$$S = OA + OB = \left(4 + \frac{9}{m}\right) + (9 + 4m) = 13 + 4m + \frac{9}{m}$$

Minimization:

For $m>0$,

$$4m + \frac{9}{m} \ge 2\sqrt{4m \cdot \frac{9}{m}} = 12$$

Equality holds when

$$4m = \frac{9}{m} \Rightarrow m = \frac{3}{2}$$

Thus,

$$S_{\min} = 13 + 12 = 25$$ $$\boxed{25}$$