Question

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?

Answer 1

`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.!