Question

Find the equation of the tangent and the normal to the curve $16x^2+9y^2=145$ at the point $(x_1,y_1)$, where $x_1=2$ and $y_1>0$.

OR

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

Answer 1

Let $P(x_1,y_1)$ where $x_1=2$ and $y_1>0$

Given curve is $16x^2+9y^2=145$.....(i)

Clearly P shall satisfy (i) so, $64+9y_1^2=145\Rightarrow y_1=3$ (as $y_1>0$

Therefore , point of contact is P(2,3).

Now $32x+18y\times \frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{16}{9y}{\begin{array}{c}\therefore \frac{dy}{dx}\end{array}]}_{\text{at P}}=-\frac{16\times 2}{9\times 3}=\frac{32}{27}=m_T$

Eq. of tangent : $y-3=-\frac{32}{27}(x-2)\Rightarrow 32x+27y=145$

Eq. of normal : $y-3=-\frac{27}{32}(x-2)\Rightarrow 27x+32y+42=0$

OR Here $f\left(x\right)=\frac{x^4}{4}-x^3-5x^2+24x+12\Rightarrow f^{\prime }\left(x\right)=x^3-3x^2-10x+24=(x-2)(x-4)(x+3)For{f}^{\prime }\left(x\right)=0,(x-2)(x-4)(x+3)=0\therefore x=2,4,-3.$

$\begin{array}{ccc}\text{Interval}& \text{Sign of f'(x)}& \text{Nature of f(x)}\\ (-\infty ,-3)& \text{-ve}& \text{Strictly decreasing}\\ (-3,2)& \text{+ve}& \text{Strictly increasing}\\ (2,4)& \text{-ve}& \text{Strictly decreasing}\\ (4,\infty )& \text{+ve}& \text{Strictly increasing}\end{array}$