Question

Equation of common tangent of $y=x^2,y=-x^2+4x-4$ is

• A. $2x-y-1=0$
• B. $y=0$
• C. $y=-4\left(x-1\right)$
• D. $y=-30x-50$

Answer 1

The correct option is A $2x-y-1=0$

Given that,

$y=x^2$ ……(1)

$y=-x^2+4x+4$ ……(1)

Differentiate equation (1) and (2) w. r. to x, we get

$\frac{dy}{dx}=2x$ …..(3)

$\frac{dy}{dx}=-2x+4$ …..(4)

Given that equation (1) and (2) have common tangent

That’s why

Slope of equation (1) =slope of equation (2)

Hence ,

$2x=-2x+4$

$or4x=4$

$x=1$

Put x=1 in equation (1) we get ,

$Y=\pm 1$

Take $y=1$

Put $x=1$, and $y=1$ in equation (3), we get

$\frac{dy}{dx}=2$

Equation of common tangent of equation (1)and (2)

$y-1=\frac{dy}{dx}(x-1)$

$y-1=2(x-1)$

$2x-y-1=0$

This is required equation