The equation of a curve is y=4/(2x+1) Given that the line 2y=x+c is a normal to the curve find the possible values of the constant c?

1 Answer
Nov 11, 2017

c in {-5/2,7/2}.

Explanation:

If we rewrite the equation (eqn.) of normal : y=1/2*x+c/2, we

find that, its slope is 1/2.

Since tangent (tgt.) is bot to normal, this means that, the

slope of tgt. is -2.

But, we know that, the slope of tgt. at any point (pt.) (x,y) on the

curve is given by, dy/dx.

For the Curve : y=4/(2x+1), dy/dx=4{-1/(2x+1)^2}d/dx(2x+1), i.e.,

dy/dx=-8/(2x+1)^2.

Therefore, dy/dx=-2 rArr -8/(2x+1)^2=-2.

:. (2x+1)^2=4, i.e., x=1/2 or x=-3/2.

From y=4/(2x+1), we have, x=1/2,y=2; &, x=-3/2,y=-2.

This means that the pts. of contact of tgts. are, (1/2,2), &, (-3/2,-2).

But pts. of contact also lie on the normal, and their co-ordinates

must satisfy the eqn. of normal.

For (1/2,2) in {(x,y) : 2y=x+c} rArr 4=1/2+c rArr c=7/2.

For (-3/2,-2), c=-4+3/2=-5/2.

To sum up, c in {-5/2,7/2}.

Enjoy Maths.!