The given equation of the curve is,
$16{\mathit{x}}^2+9{\mathit{y}}^2$ = 145 ... (i)
Let $\left({\mathit{x}}_1,{\mathit{y}}_1\right)$ lies on the curve
⇒ $16{\mathit{x}}_1^2+9{\mathit{y}}_1^2$ = 145
⇒ 16 × $2^2$ + $9{\mathit{y}}_1^2$ = 145 (Since ${\mathit{x}}_1$ = 2)
⇒ $9{\mathit{y}}_1^2$ = 145 - 64
⇒ $9{\mathit{y}}_1^2$ = 81
⇒ ${\mathit{y}}_1^2$ = 9
⇒ ${\mathit{y}}_1$ = 3 ... (${\mathit{y}}_1$ > 0)
Coordinates of the given point are (2 , 3)
$16{\mathit{x}}^2+9{\mathit{y}}^2$ = 145
Differentiating with respect to x
⇒ 32x + 18y $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = 0
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\frac{-16\mathit{x}}{9\mathit{y}}$
⇒ ${\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}_{\left(2,3\right)}$ = $\frac{-32}{27}$
Equation of tangent at (2 , 3)
y - 3 = $\frac{-32}{27}$ (x - 2)
⇒ 32x + 27y - 145 = 0
Equation of normal at (2, 3)
y - 3 = $\frac{27}{32}$ (x - 2)
⇒ 27x - 32y + 42 = 0
OR
Given that f (x) = $\frac{{\mathit{x}}^4}{4}-{\mathit{x}}^3-5{\mathit{x}}^2$ + 24x + 12
a) Strictly increasing
f (x) = $\frac{{\mathit{x}}^4}{4}-{\mathit{x}}^3-5{\mathit{x}}^2$ + 24x + 12
⇒ f' (x) = ${\mathit{x}}^3-3{\mathit{x}}^2$ - 10x + 24
Function is increasing when f ' (x) > 0
⇒ ${\mathit{x}}^3-3{\mathit{x}}^2$ - 10x + 24 > 0
⇒ (x - 4) (x + 3) (x - 2) > 0
⇒ x = - 3 , 2 , 4
x ∊ (- 3 , 2) ∪ (4 , ∞)
b) Strictly decreasing
f' (x) = 0
⇒ ${\mathit{x}}^3-3{\mathit{x}}^2$ - 10x + 24 = 0
⇒ (x - 4) (x + 3) (x - 2) < 0
⇒ x ∊ (- ∞ , - 3) ∪ (2 , 4)