The equation of the tangent to the parabola y | vedclass

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(D) Given,the equation of the parabola is $y^{2}=4x$.Comparing with $y^{2}=4ax$,we get $a=1$.The equation of the tangent to the parabola in slope form

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$x-y+1=0$

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(D) Given,the equation of the parabola is $y^{2}=4x$.Comparing with $y^{2}=4ax$,we get $a=1$.The equation of the tangent to the parabola in slope form is $y=mx+\frac{a}{m}$.Substituting $a=1$,we get $y=mx+\frac{1}{m} \quad (i)$.The tangent is inclined at an angle of $\frac{\pi}{4}$ to the positive direction of the $x$-axis,so the slope $m = 	an(\frac{\pi}{4}) = 1$.Substituting $m=1$ into equation $(i)$,we get $y = (1)x + \frac{1}{1}$.This simplifies to $y = x + 1$,or $x - y + 1 = 0$.

The equation of the tangent to the parabola $y^{2}=4x$ inclined at an angle of $\frac{\pi}{4}$ to the positive direction of $x$-axis is:

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