The point at which the line y = mx + c touche | vedclass

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(A) For the line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$,the condition is $c = \frac{a}{m}$.Substituting $c = \frac{a}{m}$ into the lin

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$\left( \frac{a}{m^2}, \frac{2a}{m} \right)$

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(A) For the line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$,the condition is $c = \frac{a}{m}$.Substituting $c = \frac{a}{m}$ into the line equation,we get $y = mx + \frac{a}{m}$.To find the point of tangency,we substitute $y = mx + \frac{a}{m}$ into $y^2 = 4ax$:$(mx + \frac{a}{m})^2 = 4ax$$m^2x^2 + 2ax + \frac{a^2}{m^2} = 4ax$$m^2x^2 - 2ax + \frac{a^2}{m^2} = 0$$(mx - \frac{a}{m})^2 = 0$This gives $x = \frac{a}{m^2}$.Substituting $x = \frac{a}{m^2}$ into $y = mx + \frac{a}{m}$,we get $y = m(\frac{a}{m^2}) + \frac{a}{m} = \frac{a}{m} + \frac{a}{m} = \frac{2a}{m}$.Thus,the point of contact is $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$.

List-$I$	List-$II$
$I$. Vertex	$A$. $8$
$II$. Length of latus rectum	$B$. $(\frac{29}{10}, \frac{-38}{10})$
$III$. Directrix	$C$. $3x+4y-1=0$
$IV$. One end of the latus rectum	$D$. $(\frac{-2}{5}, \frac{-16}{5})$
	$E$. $6$

The point at which the line $y = mx + c$ touches the parabola $y^2 = 4ax$ is

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