Question

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

• A. $c=-am$
• B. $c=-a/m$
• C. $c=-a{m}^2$
• D. $c=a/m^2$

Answer 1

The correct option is C $c=-a{m}^2$

We have,

Equation of line $\Rightarrow y=mx+c----\left(1\right)$

Equation of parabola $\Rightarrow x^2=4ay----\left(2\right)$

put the value of $y$ in equation $\left(2\right)$ and we get,

$x^2=4ay$

$\Rightarrow x^2=4a(mx+c)$

$\Rightarrow x^2=4amx+4ac$

$\Rightarrow x^2-4amx-4ac=0$

let the roots equation are equal then

$D=0$

$B^2-4AC=0$

$(-4am{)}^2=4\times 1\times (-4ac)$

$16a^2{m}^2=-16ac$

$16a^2{m}^2=-16ac$

$a^2{m}^2=-ac$

$a{m}^2=-c$

$c=-a{m}^2$

Hence, this is the answer.