AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Given, line is 2x+18y=9
slope (m)=‌

−a

b

m=‌

−2

18

=−‌

1

9

Given equation of curve is
y=x3−3x..(i)
Differentiate w.r.to ' x, we get
‌

dy

dx

=3x2−3
∴ Slope of normal =‌

−1

dy dx

=‌

−1

3x2−3

Since, normal is parallel to given line
∴ Slope of line = slope of normal
‌

−1

9

=‌

−1

3(x2−1)

⇒‌‌x2−1=3⇒x=±2
Case (i) if x=2
from Eq. (i)
y=(2)3−3(2)
y=2
∴‌‌P=(2,2)
Equation of normal is y−y1=m(x−x1)
y−2=‌

−1

9

(x−2)
x+9y=20
Case (ii) If x=−2
y=(−2)3−3(−2)=−2
∴‌‌Q=(−2,−2)
Equation of Normal is y−y1=m(x−x1)
y+2=‌

−1

9

(x+2)
x+9y=−20
Required Equation of normal is x+9y=±20