find the values of p for which the curves x^2= 9p(9-y) and x^2= p(y+1) cut each other at right angles.
Answers
The given curves are x
2
=9p(9−y) .... (1)
x
2
=p(y+1) ...... (2)
Solving (1) and (2) simultaneously, we get
9p(9−y)=p(y+1)⇒81−9y=y+1
⇒10y=80⇒y=8
x
2
=9p(9−8)⇒x
2
=9p⇒x=±3
p
Therefore, given curves intersect at points (±3
p
,8) i.e., at points P(3
p
,8) and Q(−3
p
,8)
Differentiating (1) and (2) w.r.t x, we get
2x=−9p
dx
dy
⇒
dx
dy
=
9p
−2x
...... (3) and
2x=p
dx
dy
⇒
dx
dy
=
p
2x
...... (4)
Slope of tangent to the curve (1) at point P is m
1
=
9p
−2×3
p
=
3
p
−2
Slope of tangent to the curve (2) at point P is m
2
=
p
2×3
p
=
p
6
Now given curves (1) and (2) cut at right angles,
Therefore, m
1
×m
2
=−1
⇒
3
p
−2
×
p
6
=−1
⇒
p
4
=1⇒p=4
Similarly, when the curves intersect at right angles at point Q, then p=4.
Answer:
Answer is in the attachment
