KCET 2020 Math Solved Paper

2x=y2 .....(1)
2xy=K .......(2)
Solving (1) and (2), we get
(x,y)=(

K 2 3

2

,K

1

3

)
Differentiating (1) and (2) w.r.t. x
m1=

dy

dx

=

1

y

.......(3)
m2=

dy

dx

=−

y

x

...... (4)
∵ Both curves intersect each other perpendicularly
m1m2=−1⇒−

1

x

=−1⇒x=1
⇒K

2

3

=2⇒K2=8