Question:
Let $\mathrm{AD}$ and $\mathrm{BC}$ be two vertical poles at $\mathrm{A}$ and B respectively on a horizontal ground. If $\mathrm{AD}=8 \mathrm{~m}, \mathrm{BC}=11 \mathrm{~m}$ and $\mathrm{AB}=10 \mathrm{~m}$; then the distance (in meters) of a point $\mathrm{M}$ on $\mathrm{AB}$ from the point $A$ such that $M D^{2}+M C^{2}$ is minimum is.
Solution:
$\left(\mathrm{MD}^{2}+(\mathrm{MC})^{2}=\mathrm{h}^{2}+64+(\mathrm{h}-10)^{2}+121\right.$
$=2 \mathrm{~h}^{2}-20 \mathrm{~h}+64+100+121$
$=2\left(\mathrm{~h}^{2}-10 \mathrm{~h}\right)+285$
$=2(\mathrm{~h}-5)^{2}+235$
it is minimum if $\mathrm{h}=5$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.