The equation of the tangent to the parabola y | vedclass

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(B) The slope of the chord joining the points $(3, 0)$ and $(4, 1)$ is given by $m = \frac{1 - 0}{4 - 3} = 1$.Since the tangent is parallel to this ch

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$4x - 4y = 13$

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(B) The slope of the chord joining the points $(3, 0)$ and $(4, 1)$ is given by $m = \frac{1 - 0}{4 - 3} = 1$.Since the tangent is parallel to this chord,its slope must also be $1$.Given the parabola $y = (x - 3)^2$,we differentiate with respect to $x$ to find the slope of the tangent at any point $(x, y)$:$\frac{dy}{dx} = 2(x - 3)$.Setting the slope equal to $1$:$2(x - 3) = 1$ $\Rightarrow x - 3 = \frac{1}{2}$ $\Rightarrow x = 3.5 = \frac{7}{2}$.Substituting $x = \frac{7}{2}$ into the parabola equation to find the $y$-coordinate:$y = (\frac{7}{2} - 3)^2 = (\frac{1}{2})^2 = \frac{1}{4}$.The point of tangency is $(\frac{7}{2}, \frac{1}{4})$.The equation of the tangent line with slope $m = 1$ passing through $(\frac{7}{2}, \frac{1}{4})$ is:$y - \frac{1}{4} = 1(x - \frac{7}{2})$$y - \frac{1}{4} = x - \frac{14}{4}$$x - y = \frac{14}{4} - \frac{1}{4} = \frac{13}{4}$$4x - 4y = 13$.

The equation of the tangent to the parabola $y = (x - 3)^2$ parallel to the chord joining the points $(3, 0)$ and $(4, 1)$ is:

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