IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

pottermania1990

hi,

can u pl. tell me d soln 4 this sum..........wud b of fr8 help 4 me

two towns A and B are situated on d same side of straight road at a distance a and b resp. ,from the road. perpendiculars drawn from A and B meet the road at pt.s C and D resp.dist. between C and D is c.a hospital is to be built at a pt. P on the road such that distance APB is minimum,find posn. of p

can any tell me d soln of this prob????????

sidd2004

is the ans ac/(a+b) from c
if yes then i can mail the soln.

magiclko

see the attached file,

APB = sqrrt (a² + x²) + sqrrt (b² + (c-x)²)

for APB be to be minimum, the first derivative of the above expression = 0

differentitating it wrt x, nd then equating it to 0, we have

x sqrrt (b² + (c-x)²) =- (c-x) sqrrt (a² + x²)

solving the above, finally we'll have

b² / (c-x)² = a² / x²

=> x = ac/ (a+b)

pottermania1990

thank you 4 d ans. it is ac/a+b

uzairmehdi

This is a coordinate geometry prob.The ans is at a dist of ac/a+b from C.Take the rood as X-axis C as origin.A on Y-axis at (0,a) and B at (c,b) D at (c,0) take the image of B in X-axis call it as R.Join A and R and the X-intercept of line joining AR is the required position of P.

sidd2004

see the file

amar.gupta

good work magiclko

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