Find the equation of tangent and normal to the curve `x=(2a t^2)/((1+t^2)),y=(2a t^3)/((1+t^2))` at the point for which `t=1/2dot`

Explanation

`x=frac{2 a t^{2}}{1+t^{2}}` and `y=frac{2 a t^{3}}{1+t^{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}}`
and `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 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}}}=frac{6 a t^{2}+2 a t^{4}}{4 a t}`
Slope of tangent,
`m=(frac{d y}{d x})_{t=frac{1}{2}}=frac{frac{3 a}{2}+frac{a}{8}}{2 a}=frac{frac{12 a+a}{8}}{2 a}=frac{13 a}{8} times frac{1}{2 a}=frac{13}{16}`

Now, `(x_{1}, y_{1})=(frac{2 a t^{2}}{1+t^{2}}, frac{2 a t^{3}}{1+t^{2}})=(frac{frac{a}{2}}{1+frac{1}{4}}, frac{frac{a}{4}}{1+frac{1}{4}})=(frac{frac{a}{2}}{frac{5}{4}}, frac{frac{a}{4}}{frac{5}{4}})=(frac{2 a}{5}, frac{a}{5})`
Equation of tangent is ,

`y-y_{1}=m(x-x_{1}) `

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

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

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

`Rightarrow 80 y-16 a=65 x-26 a `

`Rightarrow 65 x-80 y-10 a=0 `

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