Find an equation of the straight line that is tangent to the graph of f(x) root(x+1) and parallel to x - 6y+4 = 0.
Find the slope of the tangent to the curve 1/x + 1/y = 1 at the point (2,2).
$\displaystyle
f(x) = \sqrt{x+1}
$
slope of tangent = f'(x)
$\displaystyle
f'(x) = \frac{-1}{2\sqrt{x+1}}
$
slope of tangent, which is parallel to line x - 6y+4 = 0 , is $\displaystyle \frac{1}{6}$
so,
$\displaystyle
f'(x) = \frac{-1}{2\sqrt{x+1}}=\frac{1}{6}
$
$\displaystyle
\sqrt{x+1}=-3
$
$\displaystyle
\Rightarrow x= 8
$
$\displaystyle
y = \sqrt{8+1}=3
$
so, (8, 3) is the point on the curve where tangent is drawn.
eqn of tangent
y = mx + b
$\displaystyle
y = \frac{1}{6}x + b
$
$\displaystyle 3 = \frac{1}{6}(8) + b$
$\displaystyle
b = \frac{5}{3}
$
so, eqn of tangent is
$\displaystyle
y = \frac{1}{6}x + \frac{5}{3}
$
$\displaystyle \frac{1}{x}+\frac{1}{y}=1$
differentiating wrt x,
$\displaystyle \frac{-1}{x^2}-\frac{1}{y^2}y'=0$
$\displaystyle y' =\frac{-y^2}{x^2}$
$\displaystyle y' \;at\; (2, 2) =\frac{-(2)^2}{(2)^2}=-1$
so, slope of tangent at (2, 2) is -1
eqn of tangent
y = mx + b
2 = -1(2) + b
b = 4
eqn of tangent is
y = -x + 4