COMEDK Exam 2016 Solved Paper

Given,equation of curve is
y=x2−x+4
Slope of tangent at P(1,4) is
‌

dy

dx

=2x−1⇒(‌

dy

dx

)(1,4)=2−1=1
So, equation of tangent is
⇒‌‌y−4=1(x−1)
‌y−x=3 . . . (i)
and equation of normal at point P(1,4) is
‌y−4=−1(x−1)
‌x+y=5 . . . (ii)
Since, the tangent cuts X-axis at A.
Therefore, coordinates of A are (−3,0) and the normal cuts X-axis at B and coordinates of B are (5,0)
Therefore, area of △PAB
‌=‌

1

2

|

1	4	1
−3	0	1
5	0	1

|
‌=‌

1

2

[1(0)−4(−3−5)+1(0)]
‌=‌

1

2

|32|=16‌ sq. units ‌