The equation of a common tangent to the parab | vedclass

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(B) Let the equation of the tangent to the parabola $y = x^{2}$ be $y = mx - \frac{m^{2}}{4}$.Since this line is also tangent to $y = -(x - 2)^{2}$,we

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$y = 4(x - 1)$

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(B) Let the equation of the tangent to the parabola $y = x^{2}$ be $y = mx - \frac{m^{2}}{4}$.Since this line is also tangent to $y = -(x - 2)^{2}$,we substitute $y$ in the second equation:$mx - \frac{m^{2}}{4} = -(x - 2)^{2}$$mx - \frac{m^{2}}{4} = -(x^{2} - 4x + 4)$$x^{2} + x(m - 4) + 4 - \frac{m^{2}}{4} = 0$For the line to be a tangent,the discriminant $D$ must be zero:$D = (m - 4)^{2} - 4(1)(4 - \frac{m^{2}}{4}) = 0$$m^{2} - 8m + 16 - 16 + m^{2} = 0$$2m^{2} - 8m = 0$$2m(m - 4) = 0$Thus,$m = 0$ or $m = 4$.For $m = 4$,the tangent equation is $y = 4x - \frac{4^{2}}{4} = 4x - 4 = 4(x - 1)$.

The equation of a common tangent to the parabolas $y = x^{2}$ and $y = -(x - 2)^{2}$ is:

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