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.!