The given equation of the curve is,
16x2+9y2 = 145 ... (i)
Let (x1,y1) lies on the curve
⇒ 16x12+9y12 = 145
⇒ 16 × 22 + 9y12 = 145 (Since x1 = 2)
⇒ 9y12 = 145 - 64
⇒ 9y12 = 81
⇒ y12 = 9
⇒ y1 = 3 ... (y1 > 0)
Coordinates of the given point are (2 , 3)
16x2+9y2 = 145
Differentiating with respect to x
⇒ 32x + 18y

dy

dx

= 0
⇒

dy

dx

=

−16x

9y

⇒ (

dy

dx

)(2,3) =

−32

27

Equation of tangent at (2 , 3)
y - 3 =

−32

27

(x - 2)
⇒ 32x + 27y - 145 = 0
Equation of normal at (2, 3)
y - 3 =

27

32

(x - 2)
⇒ 27x - 32y + 42 = 0
OR
Given that f (x) =

x4

4

−x3−5x2 + 24x + 12
a) Strictly increasing
f (x) =

x4

4

−x3−5x2 + 24x + 12
⇒ f' (x) = x3−3x2 - 10x + 24
Function is increasing when f ' (x) > 0
⇒ x3−3x2 - 10x + 24 > 0
⇒ (x - 4) (x + 3) (x - 2) > 0
⇒ x = - 3 , 2 , 4
x ∊ (- 3 , 2) ∪ (4 , ∞)
b) Strictly decreasing
f' (x) = 0
⇒ x3−3x2 - 10x + 24 = 0
⇒ (x - 4) (x + 3) (x - 2) < 0
⇒ x ∊ (- ∞ , - 3) ∪ (2 , 4)