Question

What are the tangent and normal to the curve $x=\frac{2a{t}^2}{1+t^2}$, $y=\frac{2a{t}^3}{a+t^2}$ at the point for which $t=\frac{1}{2}$

• A. $16x+13y=9a$ , $13x-16y=2a$
• B. $13x-16y=2a$ , $16x+13y=9a$
• C. $16x-13y=9a$ , $13x+16y=2a$
• D. $13x+16y=2a$ , $16x-13y=9a$

Answer 1

Answer: (B)

$13x-16y=2a$ , $16x+13y=9a$

Given $x=\frac{2a{t}^2}{1+t^2}$, $y=\frac{2a{t}^3}{1+t^2}$
At $t=\frac{1}{2}$ we get
$x=\frac{2a}{5}$ , $y=\frac{a}{5}$
Hence, the point is $(\frac{2a}{5},\frac{a}{5})$
Now, $\frac{dx}{dt}=2a\frac{(1+t^2)2t-{2t}^3}{{(1+t^2)}^2}=\frac{4at}{{(1+t^2)}^2}$
$\frac{dy}{dt}=2a\frac{(1+t^2)3t^2-t^3\left(2t\right)}{{(1+t^2)}^2}=\frac{2a(3t^2+t^4)}{{(1+t^2)}^2}$
Now, $\frac{dy}{dx}=\frac{(3t^2+t^4)}{2t}$
Slope of tangent at $t=\frac{1}{2}$ is $\frac{13}{16a}$
Equation of tangent at $(\frac{2a}{5},\frac{a}{5})$ is
$y-\frac{a}{5}=\frac{13}{16a}(x-\frac{2a}{5})$
$\Rightarrow 13x-16y=2a$
Equation of normal at $(\frac{2a}{5},\frac{a}{5})$ is
$y-\frac{a}{5}=-\frac{16a}{13}(x-\frac{2a}{5})$
$\Rightarrow 16x+13y=9a$