The line y = mx + c touches the parabola y^2 | vedclass

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(D) The equation of the parabola is $y^2 = 4a(x + a)$.Let $X = x + a$,then $x = X - a$. Substituting this into the equation,we get $y^2 = 4aX$.The con

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$c = am + \frac{a}{m}$

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(D) The equation of the parabola is $y^2 = 4a(x + a)$.Let $X = x + a$,then $x = X - a$. Substituting this into the equation,we get $y^2 = 4aX$.The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4aX$ is $c' = \frac{a}{m}$,where $c'$ is the intercept on the $Y$-axis.Substituting $x = X - a$ into $y = mx + c$,we get $y = m(X - a) + c = mX + (c - am)$.Comparing this with $y = mX + c'$,we have $c' = c - am$.Equating the two expressions for $c'$,we get $c - am = \frac{a}{m}$.Therefore,$c = am + \frac{a}{m}$.

The line $y = mx + c$ touches the parabola $y^2 = 4a(x + a)$ if...

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