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

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Question

(C) Given the line $y = mx + c$ and the parabola $x^2 = 4ay$. Substituting $y$ from the line equation into the parabola equation: $x^2 = 4a(mx + c)$ $

Vedclass · Accepted Answer

$c = -am^2$

Vedclass · Answer

(C) Given the line $y = mx + c$ and the parabola $x^2 = 4ay$. Substituting $y$ from the line equation into the parabola equation: $x^2 = 4a(mx + c)$ $x^2 - 4amx - 4ac = 0$ Since the line touches the parabola,the discriminant of this quadratic equation must be zero $(D = 0)$: $B^2 - 4AC = 0$ $(-4am)^2 - 4(1)(-4ac) = 0$ $16a^2m^2 + 16ac = 0$ Dividing by $16a$ (assuming $a 
eq 0$): $am^2 + c = 0$ $c = -am^2$

The line $y = mx + c$ touches the parabola $x^2 = 4ay$ if:

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