COMEDK Exam 2017 Solved Paper

Given curve, y=‌

x3

4−x

. . . (i)
is the given curve
Differentiating Eq. (i) w.r.t. x, we get
‌

dy

dx

‌=‌

(4−x)(3x2)−x3(−1)

(4−x)2

=‌

12x2−2x3

(4−x)2

⇒(‌

dy

dx

)x=2‌=‌

12(2)2−2(2)3

(4−2)2

=‌

32

4

=8
⇒ The slope of tangent at (2,4) is 8 .
∴ The equation of tangent is
y−4=8(x−2)
⇒‌‌8x−y−12=0⇒‌ slope ‌=‌

−8

−1

=8
Now the slope of normal at P(2,4) is ‌

−1

8

.
∴ The equation of normal is
‌y−4=‌

−1

8

(x−2)
⇒‌‌x+8y−34=0