KCET 2015 Math Solved Paper

Given equation of curve,
y=x2−‌

1

x2

⇒y=x2−x−2
Differentiating both the sides, we get
y′=2x−(−2)x−3=2x+‌

2

x3

At (−1,0), we get m1=−2−2=−4
So, slope of the curve is −4
We know that, slopes of two perpendicular lines is given by
m1m2=−1
⇒m2=‌

1

4

Hence, the slope of the normal to the curve is

1

4