Question

How do you find the equation of the line that is tangent to `f(x)=1/sqrtx` and parallel to the line `x+2y-6=0`?

Answer 1

The equation is `x + 2y -3 = 0`

Explanation:

A parallel line will have an equal slope. The slope of `x + 2y - 6= 0` can be obtained by converting to `y = mx + b`.

`2y = 6 - x`

`y = 3 - 1/2x`

The slope, `m` in `y = mx+ b`, is `-1/2`. Since the derivative represents the gradient at any point in the function's domain, we should find the derivative.

`f(x) = 1/sqrt(x)`

`f(x) = 1/x^(1/2)`

`f(x) = x^(-1/2)`

`f'(x) = -1/2x^(-3/2)`

`f'(x) = -1/(2xsqrt(x)`

We set this to `-1/2` and solve for `x`.

`-1/2 = -1/(2xsqrt(x))`

`-2xsqrt(x) = -2`

`xsqrt(x) = -2/-2`

`xsqrt(x) = 1`

`(xsqrt(x))^2 = 1^2`

`x^2(x) = 1`

`x^3 = 1`

`x =1`

We know now that the tangent line that is parallel to `x + 2y - 6= 0` makes contact with the curve at `x = 1`. The complete point `(x, y)` is:

`f(1) = 1/sqrt(1) = 1/1 = 1`

So, the point of tangency is `(1, 1)` and the slope is `-1/2`. We can derive an equation for this tangent now:

`y - y_1 = m(x- x_1)`

`y - 1 = -1/2(x - 1)`

`y - 1 = -1/2x + 1/2`

`y = -1/2x + 3/2`

Let's finally convert into the form `Ax + By - C = 0`, just as it was written in the question.

`1/2x + y - 3/2 = 0`

Multiply by a factor of `2` so that all the coefficients are integers.

`x + 2y - 3 = 0`

Hopefully this helps!