Question

The slope of normal to the curve x = `sqrt"t"` and y = `"t" - 1/sqrt"t"`at t = 4 is _____.

विकल्प

• `(- 17)/4`

• `4/17`

• `(- 4)/17`

• `17/4`

Answer 1

The slope of normal to the curve x = `sqrt"t"` and y = `"t" - 1/sqrt"t"`at t = 4 is `underline((- 4)/17)`.

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/(2"t"^(3//2)))`

`= (2"t"^(3//2) + 1)/"t"`

At t = 4, `"dy"/"dx" = (2(4)^(3//2) + 1)/4 = 17/4`

∴ Slope of normal at t = 4, - `1/("dy"/"dx") = - 4/17`