Find the equations of the tangent and the normal, to the curve $16{\mathrm{x}}^2+9{\mathrm{y}}^2=145$ at the point ${\mathrm{x}}_1=2\text{ and }{\mathrm{y}}_1>0$.

OR

Find the intervals in which the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^4}{4}-{\mathrm{x}}^3-5{\mathrm{x}}^2+24\mathrm{x}+12$ is (a) strictly increasing, (b) strictly decreasing.

Given: $16{\mathrm{x}}^2+9{\mathrm{y}}^2=145$
Given that $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ lies on given equation,
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}16{\mathrm{x}}_1^2+9{\mathrm{y}}_1^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=145\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}16(2{)}^2+9{\mathrm{y}}_1^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=145\left[\because {\mathrm{x}}_1=2\text{ (given) }\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9{\mathrm{y}}_1^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=145-64=81\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3\left[\because {\mathrm{y}}_1>0\text{ (given) }\right]\end{array}$
$\therefore$ Point of contact is (2,3).
Differentiating equation (i) w.r.t. x, which will give us the slope of the tangent.
$\begin{array}{l}32\mathrm{x}+18\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{32\mathrm{x}}{18\mathrm{y}}\end{array}$
Slope of tangent at (2,3)
$\begin{array}{l}={\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{(2,3)}\\ =-\frac{32\left(2\right)}{18\left(3\right)}=-\frac{64}{54}\\ =-\frac{32}{27}\end{array}$
$\therefore$ Equation of tangent is
$\begin{array}{l}\mathrm{y}-3=\mathrm{m}(\mathrm{x}-2)\\ \Rightarrow \mathrm{y}-3=-\frac{32}{27}(\mathrm{x}-2)\\ \Rightarrow 27\mathrm{y}-81=-32\mathrm{x}+64\\ \Rightarrow 32\mathrm{x}+27\mathrm{y}=145\end{array}$
The slope of the normal =$\frac{1}{\mathrm{Slope} \mathrm{of} \mathrm{tangent}}$
$=\frac{27}{32}$
$\therefore$ Equation of normal is
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{y}-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{27}{32}(\mathrm{x}-2)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}32\mathrm{y}-96\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=27\mathrm{x}-54\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}27\mathrm{x}-32\mathrm{y}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-42\end{array}$
Therefore, the equations of the tangent and the normal are 32x+27y=145 and 27x-32y=-42
respectively.

OR

Given: $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^4}{4}-{\mathrm{x}}^3-5{\mathrm{x}}^2+24\mathrm{x}+12$
Now, differentiate both sides with respect to x, it follows
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{4{\mathrm{x}}^3}{4}-3{\mathrm{x}}^2-10\mathrm{x}+24\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2-10\mathrm{x}+24\end{array}$
For critical points, put ${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=0$
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^3-3{\mathrm{x}}^2-10\mathrm{x}+24\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}-2)\left({\mathrm{x}}^2-\mathrm{x}-12\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}-2)(\mathrm{x}-4)(\mathrm{x}+3)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=2,4,-3\end{array}$
Therefore, we have the intervals $(-\mathrm{\infty },-3),(-3,2)$
$(2,4)\text{ and }(4,\mathrm{\infty })$

f(x) is increasing in (-3,2) and$(4,\infty )$ and decreasing in$(-\infty ,-3) and (2,4)$