Question

Find the equation of tangent and normal to the curve

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

at

t = \frac{1}{2} .

Accepted Answer

Given $x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}}, t = \frac{1}{2}$

Thus, $P = (\frac{2 a {(\frac{1}{2})}^{2}}{1 + {(\frac{1}{2})}^{2}}, \frac{2 a {(\frac{1}{2})}^{3}}{1 + {(\frac{1}{2})}^{2}})$

$\Rightarrow P = (\frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}})$

$\Rightarrow P = (\frac{2 a}{5}, \frac{a}{5})$

Now, $\frac{d x}{d t} = \frac{4 a t (1 + t^{2}) - 2 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} = \frac{4 a t}{{(1 + t^{2})}^{2}}$

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

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

$\Rightarrow \frac{d y}{d x} = \frac{6 a t^{2} - 2 a t^{4}}{4 a t}$

Thus, slope $m = {(\frac{d y}{d x})}_{P} = \frac{6 a {(\frac{1}{2})}^{2} - 2 a {(\frac{1}{2})}^{4}}{4 a \times \frac{1}{2}}$

$\Rightarrow m = {(\frac{d y}{d x})}_{P} = \frac{\frac{3 a}{2} - \frac{a}{8}}{2 a} = \frac{13}{16}$

Thus, the equation of tangent is given by: $y - y_{1} = m (x - x_{1})$

$\Rightarrow y - \frac{a}{5} = \frac{13}{16} (x - \frac{2 a}{5})$

$\Rightarrow 16 y - \frac{16 a}{5} = 13 x - \frac{26 a}{5}$

$\Rightarrow 13 x - 16 y - 2 a = 0$

The equation of normal is given by $y - y_{1} = - \frac{1}{m} (x - x_{1})$

$\Rightarrow y - \frac{a}{5} = - \frac{16}{13} (x - \frac{2 a}{5})$

$\Rightarrow 16 x + 13 y - 9 a = 0$ .

R D Sharma Solutions for Chapter: Tangents and Normals, Exercise 1: EXERCISE

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Tangents and Normals, Exercise 1: EXERCISE

Questions from R D Sharma Solutions for Chapter: Tangents and Normals, Exercise 1: EXERCISE with Hints & Solutions

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