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A straight line through the point (h,k) ...

A straight line through the point `(h,k)` where `hgt0` and `kgt0,` makes positive intercepts on the coordinate axes. Then the minimum length of line intercepted between the coordinate axes is

A

`(h^(2//3)+k^(2//3))^(3//2)`

B

`(h^(3//2)+k^(3//2))^(2//3)`

C

`(h^(2//3)-k^(2//3))^(2//3)`

D

`(h^(3//2)-k^(3//2))^(2//3)`

Text Solution

Verified by Experts

The correct Answer is:
A

Let the straight line be
`y-k=m(x-h), " where " mlt0`
It meets X and Y axes at `A(h-k//m,0) and B(0,k-mh)`
`thereforeAB^2-(h-k/m)^2+(k-mh)^2`
`rArr d/(dn)(AB^2)=2h-k/mk/m^2-2h(k-mh)`
`rArrd/(dx)(AB)^2=2{(hk)/m^2-k^2/m^3-hk+mh^2}`
`rArrd/(dx)(Ab)^2=2/m^3{(mhk-k^2-m^3 " hk " +m^4 h^2}`
`rArr d/(dx)(AB)^2=2/m^3{(mh-k)(m^3 h+k)}`
`d/(dx)(AB)^2=0=rArrm=k/h or,m=-(k/h)^(1//3)`
`" since " mlt0," therefore " m=-(k/h)^(1/3)`
it can be easily checked that `d^2/(dm^2)(AB)^2` is positive for this value of m.thus,`Ab^2`is minimumfor `m=-(k//h)^(1//3)`
Putting `m=-(k//h)^(1//3)` in (i) we get
`(AB)^2 =(h^(2//3)+k^(2//3))^3 "  " AB=h^(2//3)+k^(2//3))^(3//2)`
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