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(C) The equation of a tangent to the parabola $y = x^2$ with slope $m$ is given by $y = mx - \frac{m^2}{4} \dots (1)$.The equation of the parabola $y

Vedclass · Accepted Answer

$y = 4(x - 1)$

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(C) The equation of a tangent to the parabola $y = x^2$ with slope $m$ is given by $y = mx - \frac{m^2}{4} \dots (1)$.The equation of the parabola $y = -(x - 2)^2$ can be written as $(x - 2)^2 = -y$. The equation of a tangent to this parabola with slope $m$ is $y - k = m(x - h) + \frac{a}{m}$,where the vertex is $(h, k) = (2, 0)$ and $4a = -1 \implies a = -1/4$.Thus,the tangent equation is $y = m(x - 2) + \frac{-1/4}{m} = mx - 2m - \frac{1}{4m} \dots (2)$.For the lines to be the same,the intercepts must be equal: $-\frac{m^2}{4} = -2m - \frac{1}{4m}$.Multiplying by $-4m$,we get $m^3 = 8m^2 + 1 \implies m^3 - 8m^2 - 1 = 0$. This approach is complex; let us use the condition for common tangents.Alternatively,for $y = x^2$ and $y = -(x-2)^2$,the common tangents are $y = 0$ and $y = 4(x - 1)$.

Find the equation of the common tangent to the parabolas $y = x^2$ and $y = -(x - 2)^2$.

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