Question

Let $P\equiv (-1,0),S_1\equiv x^3-y-3x^2-8x-4=0,S_2\equiv 3x^2-y+7x+4=0.$
Which of the following is/are true:

Answer 1

For $S_1=0,\frac{dy}{dx}=3x^2-6x-8$
For $S_2=0,\frac{dy}{dx}=6x+7$
Solving $S_1=0$ and $S_2=0$ we get $x=-1,-1,8$
at $x=-1,m_1=m_2=1$
$\Rightarrow S_1=0,S_2=0$ touch each other
Equation of common normal at $P,$
$y-0=-1(x+1)$ $\Rightarrow x+y+1=0$
At $x=8,m_1=136,m_2=55\Rightarrow S_1=0,S_2=0$ intersect each other
Equation of common tangent at $P,$
$y-0=1(x+1)$
$\Rightarrow y=x+1$