Question

Find the point at which tangent to the curve y=sqrt(1+2x) is perpendicular to the line 6x+2y=1

Answer 1

Step-by-step explanation:

Curve:

$y = \sqrt{1 + 2x}$

$\implies \frac{dy}{dx} = \frac{2}{2 \sqrt{1 + 2x} }$

$\implies \frac{dy}{dx}_{(x = h \: \: y = k)} = \frac{1}{ \sqrt{1 + 2h} }$

Now, the tangent of the curve is perpendicular to line 6x + 2y = 1

So,

$\frac{1}{ \sqrt{1 + 2h} } . \frac{ - 1}{ 3} = - 1$

$\implies3 \sqrt{1 + 2h} = 1$

$\implies9(1 + 2h) = 1$

$\implies1 + 2h = \frac{1}{9}$

$\implies2h = - \frac{8}{9}$

$\implies \: h = - \frac{4}{9}$

$\therefore \: k = \frac{1}{3}$

Hence, the required point $(\frac{1}{3},-\frac{4}{9})$