1.

Find the value of p for which curves `x^2=9p(9-y)` and `x^2=p(y+1)` cut each other at right angles.

Answer» Equations of the given curves,
`x^2 = 9p(9-y)->(1)`
`x^2 = p(y+1)->(2)`
From (1) and (2),
`9p(9-y) = p(y+1)`
`=>81p-9py = py+p`
`=>10py = 80p=> y =8`
`:. x^2 = (8+1)p = 9p`
`=>x = 3sqrtp`
So, point of intersection of these curve is `(3sqrtp,8)`.
Now, we have to calculate their slopes by differentiating their equations.
For first curve,
Differentiating its equation,
`2x = 9p(-dy/dx)=>dy/dx = -(2x)/(9p)`
At point `(3sqrtp, 8)`, `dy/dx = m_1 = -(2sqrtp)/(3p)`
For second curve,
Differentiating its equation,
`2x = p(dy/dx) =>dy/dx = (2x)/p`
At point `(3sqrtp, 8)`, `dy/dx = m_2 =(6sqrtp)/(p)`
As these two curves are at right angles, multiplication of their slopes should be equal to `-1`.
`:. -(2sqrtp)/(3p)**(6sqrtp)/(p) = -1`
`=>-12 = -3p => p = 4`


Discussion

No Comment Found

Related InterviewSolutions