If the tangent to the curve `y = 1 – x^(2) at x = alpha`, where` 0 lt alpha lt 1,` meets the axes at P and Q. Also `alpha` varies, the minimum value of the area of the triangle OPQ is k times area bounded by the axes and the part of the curve for which `0 lt x lt 1`, then k is equal to

Answer

2

Explanation

`A_(1)=` area under the curve and axes
`=int_(0)^(1)(1-x^92))dx=[x-(x^(3))/(3)]_(0)^(1)-(2)/(3)` .(i)
now `y=1-x^(2)`
`becausey^(')=-2x`
`becausey^(')(alpha,1-alpha^(2))=-2alpha`
Equation of tangent to the curve `y=1-x^(2)`
`implies(y-(1-alpha^(2)))=-2alpha(x-alpha)`
`implies2alphax+y=alpha^(2)+1`
`becausep-=((alpha^(2)+1)/(2alpha),0),Q-=(0,alpha^(2)+1)`
`A=` area of Delta `POQ=(1)/(2)(OP)(OQ)`
`=(1)/(4)((alpha^(2)+1)^(2))/(alpha)`
`because(dA)/(dalpha)=(1)/(4)(alpha2(alpha^(2)+1)2alpha^(2)+1)^(2))/(alpha^(2))=(1)/(4)(3alpha^(4)+2alpha^(2)-1)/(alpha^(2))`
For maximum/minimum
`(dA)/(dalpha)=0`
`implies3alpha^(4)+2alpha^(2)-1=0alpha^(2)=(1)/(3)`
`impliesalpha=+-(1)/(sqrt(3))`
Since `0ltalphalt1impliesalpha-(1)/(sqrt(3)),(d^(2)A)/(dalphal^(2))gt0`
Hence A is minimum
`A=(1)/(4)(((1)/(3)+1)^(2))/((1)/(sqrt(3)))=(4)/(3sqrt(3))` ..(ii)
Since `A=kA_(1)` (given)
`implies(4)/(3sqrt(3))=(2)/(3)kimpliesk=(2)/(sqrt(3))`