Basic maths-calculus-Q7

+1 vote

The point on the curve y = \sqrt x which is closest to the point (4,0) is

(A) (4,2)

(B) \left ( \frac{7}{2} , \sqrt{\frac{7}{2}} \right)

(C) \left ( \frac{1}{2} , {\frac{1}{ \sqrt2}} \right)

(D) \left ( \frac{1}{2} , {\frac{3}{ \sqrt2}} \right)

asked May 19 in Basic Maths by gbeditor (44,490 points)
reshown May 24 by gbeditor

2 Answers

0 votes
 
Best answer

Distance of (4,0) from any point (x,y) is \sqrt{(x-4)^2 + (y-0)^2}

From the above options, only A,B,C points are the part of curve , now find distance from (4,0) to these options A, B, C . hence correct answer is B.

 

answered May 27 by deepak-gatebook (112,390 points)
can you please tell formal solution apart from substitution and elimination method
wait I will give you solution. The line from the point(4,0) to the given curve must be a normal to the curve as perpendicular line have the shortest length. So we will use the same concept here. (m1m2 = -1)
+1 vote

This might help with the solution

answered May 28 by tsabhineet21 (7,650 points)
Answer:
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