The slopee of normal to the curve `x=sqrt(t) and y=t-(1)/(sqrt(t))` at t=4 is . . .

Options

A. `(-17)/(4)`
B. `(4)/(17)`
C. `(-4)/(17)`
D. `(17)/(4)`

Answer

C

Explanation

We have, `x=sqrt(t) and y=t-(1)/(sqrt(t))`
Now, `(dx)/(dt)=(1)/(2sqrt(t)),(dy)/(dt)=1+(1)/(2)t^(-3//2)`
`therefore(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(1+(1)/(2)t^(-3//2))/((1)/(2sqrt(t)))=2sqrt(t)(1+(1)/(2)t^(-3//2))`
`=2sqrt(t)(1+(1)/(2t^(3//2)))=(2t^(3//2)+1)/(t)`
At `t=4,(dy)/(dx)=(2(4)^(3//2)+1)/(4)=(17)/(4)`
`therefore` Slope of normal at `t=4,-(1)/(((dy)/(dx)))=-(4)/(17)`