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A curve 'C' with negative slope through ...

A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that `PQ=1`. Then
The orthogonal trajectories of 'C' are

A

Circles

B

Parabolas

C

Ellipses

D

Hyperbolas

Text Solution

Verified by Experts

The correct Answer is:
A
Replacing `(dy)/(dx) by (-dy)/(dx)` in (1), we get
`(-dx)/(dy) = (-y)/sqrt(1-y^(2))`
`rArr -2intdx= int(-2ydy)/sqrt(1-y^(2))`
`rArr -2intdx = int(-2ydy)/sqrt(1-y^(2))`
`rArr -2x-2c=2sqrt(1-y^(2))`
`rArr (x+c)^(2)+y^(2)=1`, which are required orthogonal trajectories.
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