JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 2)
Let a variable line of slope $$m>0$$ passing through the point $$(4,-9)$$ intersect the coordinate axes at the points $$A$$ and $$B$$. The minimum value of the sum of the distances of $$A$$ and $$B$$ from the origin is
30
15
10
25
Explanation
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$$\begin{aligned} & L:(y+9)=m(x-4) \\ & A:\left(4+\frac{9}{m}, 0\right) \\ & B:(0,-9-4 m) \\ & O A+O B]_{\min } \\ & \Rightarrow E_{\min }=4+\frac{9}{m}+9+4 m \\ & E=13+\frac{9}{m}+4 m \\ & \frac{d E}{d M}=0 \Rightarrow-\frac{9}{m^2}+4=0 \Rightarrow m= \pm \frac{3}{2} \end{aligned}$$
$$\begin{aligned} & \frac{d^2 E}{d M^2}=\frac{18}{m^3}>0 \text { for } \quad m=\frac{3}{2} \\ & \therefore E_{\min }=4+6+9+6=25 \end{aligned}$$
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