If equation of normal to the curve $x=\sqrt{t}$, $y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}$ at…
If equation of normal to the curve $x=\sqrt{t}$, $y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}$ at $\mathrm{t}=4$ is
$8 x+2 y=23$
$34 x-8 y=40$
$8 x+6 y=37$
$8 x+34 y=135$
Solution
$x=\sqrt{\mathrm{t}}, y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}$
$\therefore \quad y=x^2-\frac{1}{x}$, at $\mathrm{t}=4, x=2$ and $y=\frac{7}{2}$
$\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x+\frac{1}{x^2}$
$\therefore \quad$ Slope of the normal at $\mathrm{t}=4$ is $\frac{-1}{\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{\mathrm{t}=4}}=\frac{-4}{17}$
$\therefore \quad$ Required equation is $\left(y-\frac{7}{2}\right)=\frac{-4}{17}(x-2)$ i.e., $8 x+34 y=135$