Question

If the line $y=mx+c$ touches the parabola $y^2=12\left(x+3\right)$ exactly for one value of $m$ $\left(m>0\right)$ , then the value of $\frac{c+m}{c-m}$ is equal to

Answer 1

Answer: 1.4

Equation of the tangent to $y^2=12\left(x+3\right)$ with slope $m$ is
$y=m\left(x+3\right)+\frac{3}{m},$ which is same as $y=mx+c$
$\Rightarrow c=3m+\frac{3}{m}$
$\Rightarrow 3m^2-mc+3=0$ which is quadratic in $‘m’$
Since, it has both roots equal
$\Rightarrow c^2-36=0\Rightarrow c=\pm 6$
$\Rightarrow c=6\left\{m>0\Rightarrow c>0\right\}$
$\Rightarrow m=1\&c=6$
$\Rightarrow \frac{c+m}{c-m}=\frac{7}{5}=1.4$