Find the equation of the tangent and the normal to the following curve at the indicated point.
$x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}}$ at $t = 1 / 2$ .

Open in App

Solution

$x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}}$
at $t = 1 / 2$
$\frac{d x}{d t} = \frac{(1 + t^{2}) 4 a t - 2 a t^{2} (2 t)}{(1 + t^{2})^{2}} = \frac{4 a t}{(1 + t^{2})^{2}}$

$\frac{d y}{d t} = \frac{(1 + t^{2}) 6 a t^{2} - 2 a t^{3} (2 t)}{(1 + t^{2})^{2}} = \frac{6 a t^{2} + 2 a t^{4}}{(1 + t^{2})^{2}}$

$\frac{d y}{d t} \times \frac{d t}{d x} = \frac{6 a t^{2} + 2 a t^{4}}{(1 + t^{2})^{2}} \times \frac{(1 + t^{2})^{2}}{4 a t}$

$\frac{d y}{d x} = \frac{3 t + t^{3}}{2}$
at point $t = \frac{1}{2}$
$\frac{d y}{d x} = \frac{13}{16}$
Slope of tangent $= \frac{13}{16}$
Equation of tangent
$y - y_{1} = m (x - x_{1})$
$\Rightarrow y_{1} = \frac{2 a (1 / 2)^{3}}{1 + {(\frac{1}{2})}^{2}} = \frac{a}{5}, x_{1} = \frac{2 a {(\frac{1}{2})}^{2}}{1 + {(\frac{1}{2})}^{2}} = \frac{2 a}{5}$

$\Rightarrow y - \frac{a}{5} = \frac{13}{16} (x - \frac{2 a}{5})$
Slope of normal $= \frac{- 16}{13}$
Equation of normal
$y - \frac{a}{5} = \frac{- 16}{13} (x - \frac{2 a}{5})$ .

Suggest Corrections

Similar questions

x = \frac{2 a t^{2}}{1 + t^{2}}

y = \frac{2 a t^{3}}{a + t^{2}}

t = \frac{1}{2}

y = 2 x^{2} - 3 x - 1

(1, - 2)

y = x^{2} + 4 x + 1

(- 1, - 2)

x = a sec t, y = b tan t

t

x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = 1 / 2

Join BYJU'S Learning Program