The equation of the tangent to the parabola y | vedclass

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(B) The given parabola is $y^2 = 4(x + \frac{5}{4})$.Comparing this with $Y^2 = 4aX$,we have $a = 1$,$Y = y$,and $X = x + \frac{5}{4}$.The slope of th

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$2x - y + 3 = 0$

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(B) The given parabola is $y^2 = 4(x + \frac{5}{4})$.Comparing this with $Y^2 = 4aX$,we have $a = 1$,$Y = y$,and $X = x + \frac{5}{4}$.The slope of the line $y = 2x + 7$ is $m = 2$.The equation of a tangent to the parabola $Y^2 = 4aX$ with slope $m$ is $Y = mX + \frac{a}{m}$.Substituting the values,we get $y = 2(x + \frac{5}{4}) + \frac{1}{2}$.$y = 2x + \frac{5}{2} + \frac{1}{2}$.$y = 2x + 3$.Thus,the equation of the tangent is $2x - y + 3 = 0$.

The equation of the tangent to the parabola $y^2 = 4x + 5$ parallel to the line $y = 2x + 7$ is

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