TS EAMCET 2017 Code A Solved Paper

Given equation of the curves are,
x2=8y....(1)
xy=8....(2)
On solving equation (1) and (2), we get
x=4
y=2
Therefore, the point of intersection of given curves is (4,2).
Slope of tangent to the curve (1) at point (4,2) is calculated as
x2=8y
2x=8‌‌

dy

dx

⇒‌‌

dy

dy

=‌‌

x

4

⇒m1=1
Slope of tangent to the curve ( 2 ) at point (4,2) is calculated
as
xy=8
x‌‌

dy

dx

+y=0⇒‌‌

dy

dx

=−‌‌

y

x

⇒m2=−‌‌

1

2

The angle θ between the curves is calculated as,
tan‌θ=(‌‌

m2−m1

1+m1m2

)
=(‌‌

‌‌ −1 2 −1

1−‌‌ 1 2

)
tan‌θ=−3
θ=tan−1(−3)