The equation of the common tangent to the parabola y=x2 and y=−(x−2)2 is /are

A
$\begin{aligned} y = 4 (x - 1) \end{aligned}$
B
$\begin{aligned} y = 0 \end{aligned}$
C
$y = - 4 (x - 1)$
D
$y = - 30 x - 50$

Solution:

Any point on the parabola $y = x^{2}$ is $(t, t^{2})$ .

Now tangent at $(t, t^{2})$ is

$\begin{array}{l} x x_{1} = \frac{1}{2} (y + y_{1}) \\ \Rightarrow t x = \frac{1}{2} (y + t^{2}) \\ \Rightarrow 2 t x - y - t^{2} = 0 \end{array}$

If it is a tangent to the parabola, $y = - (x - 2)^{2}$ , then

$\begin{array}{l} 2 t x - t^{2} = - (x - 2)^{2} \\ \Rightarrow 2 t x - t^{2} = - x^{2} + 4 x - 4 \\ \Rightarrow x^{2} + 2 (2 - t) x + (t^{2} - 4) = 0 \end{array}$

Since it has equal roots, so

$\begin{array}{l} \Rightarrow D = 0 \\ 4 (2 - t)^{2} - 4 (t^{2} - 4) = 0 \\ \Rightarrow (2 - t)^{2} - (t^{2} - 4) = 0 \end{array}$

$\Rightarrow t = 2$ , $0$

Hence, the equation of the common tangent is $y = 4 x - 4$ , $y = 0$ .

The equation of the common tangent to the parabola $y = x^{2}$ and $y = - (x - 2)^{2}$ is /are

Solution:

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