AP EAMCET Engineering 2013 Solved Paper

Given curves are
x2+py2=1 .......(i)
and qx2+y2=1 ......(ii)
On differentiating Eq. (i), w.r.t., x we get
2x+2yp

dy

dx

=0

dy

dx

=m1=−

x

py

⇒

dy

dx

=m1=−

x

py

On differentiating Eq. (ii), w.r.t. to x, we get
2qx+2y

dy

dx

=0
⇒

dy

dx

=m2=

−qx

y

Since, both the curves are orthogonal to each other,
Then, m1m2=−1
⇒

−x

py

.

−qx

y

=−1
⇒ qx2=−py2 ......(iii)
⇒ q(1−py2)=−py2 [from Eq. (i)]
⇒ q−pqy2=−py2
∴ y2=

q

pq−p

and x2=

−p

pq−p

On putting x2 and y2 in Eq. (ii) we get
−

pq

pq−q

+

q

pq−p

=1
⇒ −pq+q=pq−p⇒p+q=2pq
⇒

1

p

+

1

q

=2