The equation of the tangent to the parabola y | vedclass

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(C) The equation of the parabola is $y^2=16x$,so $4a=16$,which gives $a=4$. The given line is $3x-4y+5=0$,which can be written as $y=\frac{3}{4}x+\fra

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$4x+3y+9=0$

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(C) The equation of the parabola is $y^2=16x$,so $4a=16$,which gives $a=4$. The given line is $3x-4y+5=0$,which can be written as $y=\frac{3}{4}x+\frac{5}{4}$. The slope of this line is $m_1=\frac{3}{4}$. Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes m_1 = -1$. Thus,$m = -\frac{4}{3}$. The equation of a tangent to the parabola $y^2=4ax$ with slope $m$ is $y=mx+\frac{a}{m}$. Substituting $a=4$ and $m=-\frac{4}{3}$,we get $y=-\frac{4}{3}x+\frac{4}{-4/3}$. $y=-\frac{4}{3}x-3$. Multiplying by $3$,we get $3y=-4x-9$,which simplifies to $4x+3y+9=0$.

The equation of the tangent to the parabola $y^2=16x$,which is perpendicular to the line $3x-4y+5=0$,is given by

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