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: Equate the two curves to find the point of contact.
The given curves are
\[ y=x^3-3x^2-8x-4 \] and \[ y=3x^2+7x+4 \] At the point of contact \(P\), both curves have the same value of \(y\). Hence, \[ x^3-3x^2-8x-4=3x^2+7x+4 \] Simplifying, \[ x^3-6x^2-15x-8=0 \] Now factorize the cubic expression: \[ x^3-6x^2-15x-8=(x+1)(x^2-7x-8) \] Further factorization gives \[ (x+1)(x-8)(x+1)=0 \] \[ (x+1)^2(x-8)=0 \] Since the curves touch each other, the repeated root gives the point of tangency. Therefore, \[ x=-1 \]

Step 2: Find the corresponding \(y\)-coordinate.
Substitute \(x=-1\) into the second curve: \[ y=3(-1)^2+7(-1)+4 \] \[ =3-7+4 \] \[ =0 \] Thus, the point of contact is \[ P(-1,0) \]

Step 3: Find the slope of the common tangent.
Differentiate the first curve: \[ y=x^3-3x^2-8x-4 \] \[ \frac{dy}{dx}=3x^2-6x-8 \] At \(x=-1\), \[ m=3(-1)^2-6(-1)-8 \] \[ =3+6-8 \] \[ =1 \] Hence, the slope of the common tangent is \[ m=1 \]

Step 4: Use point-slope form of tangent equation.
Using the point \(P(-1,0)\) and slope \(m=1\), \[ y-0=1(x+1) \] \[ y=x+1 \] Rearranging, \[ x-y+1=0 \]

Step 5: Final Answer.
Therefore, the equation of the common tangent is \[ \boxed{x-y+1=0} \]