Question

If the curves \(y=x^3-3x^2-8x-4\) and \(y=3x^2+7x+4\) touch each other at a point \(P\), then the equation of common tangent at \(P\) is

• A. \(x-y+1=0\)
• B. \(2x-y+1=0\)
• C. \(x+y+1=0\)
• D. \(2x+y+1=0\)

Hint:

If two curves touch each other, then after equating them, the point of contact appears as a repeated root.

Answer 1

The Correct Option is A

Step 1: Set the curves equal at the contact point.
The curves $y=x^3-3x^2-8x-4$ and $y=3x^2+7x+4$ meet where $x^3-3x^2-8x-4=3x^2+7x+4$, that is $x^3-6x^2-15x-8=0$.
Step 2: Use the touching condition.
Touching means a repeated root. Factoring, $x^3-6x^2-15x-8=(x+1)^2(x-8)$, so the double root is $x=-1$, the point of contact.
Step 3: Find the $y$ coordinate.
Using the simpler parabola, $y=3(-1)^2+7(-1)+4=3-7+4=0$, so $P(-1,0)$.
Step 4: Compute the common slope.
Differentiating the cubic, $\dfrac{dy}{dx}=3x^2-6x-8$; at $x=-1$ this is $3+6-8=1$. (The parabola gives $6x+7=-6+7=1$ too, confirming the touch.)
Step 5: Write the tangent line.
Through $(-1,0)$ with slope $1$: $y-0=1\,(x+1)$, so $y=x+1$.
Step 6: Put it in standard form.
Rearranging, $x-y+1=0$.
\[ \boxed{x-y+1=0} \]