The equation of the tangent to the parabola y | vedclass

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(D) Given the parabola $y^2 = 9x$,we have $4a = 9$,so $a = \frac{9}{4}$.The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = m

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$Both (b) and (c)$

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(D) Given the parabola $y^2 = 9x$,we have $4a = 9$,so $a = \frac{9}{4}$.The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Substituting $a = \frac{9}{4}$,the equation becomes $y = mx + \frac{9}{4m}$.Since the tangent passes through the point $(4, 10)$,we substitute $x = 4$ and $y = 10$:$10 = 4m + \frac{9}{4m}$Multiply by $4m$ to get the quadratic equation:$40m = 16m^2 + 9$$16m^2 - 40m + 9 = 0$Solving for $m$ using the quadratic formula or factoring:$(4m - 9)(4m - 1) = 0$$m = \frac{9}{4}$ or $m = \frac{1}{4}$.For $m = \frac{9}{4}$,the equation is $y = \frac{9}{4}x + \frac{9}{4(9/4)}$ $\Rightarrow y = \frac{9}{4}x + 1$ $\Rightarrow 4y = 9x + 4$ $\Rightarrow 9x - 4y + 4 = 0$.For $m = \frac{1}{4}$,the equation is $y = \frac{1}{4}x + \frac{9}{4(1/4)}$ $\Rightarrow y = \frac{1}{4}x + 9$ $\Rightarrow 4y = x + 36$ $\Rightarrow x - 4y + 36 = 0$.Thus,both equations $(b)$ and $(c)$ are correct.

The equation of the tangent to the parabola $y^2 = 9x$ which passes through the point $(4, 10)$ is:

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