To determine the value of [tex]\( m \)[/tex] for which the line [tex]\( mx + y + 1 = 0 \)[/tex] is tangent to the parabola [tex]\( y^2 = -6x \)[/tex], follow these steps.
1. Express [tex]\( y \)[/tex] from the line equation:
Given the line equation:
[tex]\[
mx + y + 1 = 0
\][/tex]
solve for [tex]\( y \)[/tex]:
[tex]\[
y = -mx - 1
\][/tex]
2. Substitute [tex]\( y \)[/tex] in the parabola equation:
Given the parabola equation:
[tex]\[
y^2 = -6x
\][/tex]
substitute [tex]\( y = -mx - 1 \)[/tex] into this equation:
[tex]\[
(-mx - 1)^2 = -6x
\][/tex]
3. Expand and simplify the equation:
Expanding the left-hand side:
[tex]\[
(mx + 1)^2 = m^2x^2 + 2mx + 1
\][/tex]
So the equation becomes:
[tex]\[
m^2x^2 + 2mx + 1 = -6x
\][/tex]
4. Combine like terms:
Rearrange the equation to combine all terms on one side:
[tex]\[
m^2x^2 + (2m + 6)x + 1 = 0
\][/tex]
5. Determine when the line is tangent to the parabola:
For the line to be tangent to the parabola, the quadratic equation in [tex]\( x \)[/tex]:
[tex]\[
m^2x^2 + (2m + 6)x + 1 = 0
\][/tex]
must have exactly one solution. This happens when the discriminant of the quadratic equation is zero.
The discriminant [tex]\( \Delta \)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
For our equation, [tex]\( a = m^2 \)[/tex], [tex]\( b = 2m + 6 \)[/tex], and [tex]\( c = 1 \)[/tex]. The discriminant is:
[tex]\[
\Delta = (2m + 6)^2 - 4(m^2)(1)
\][/tex]
6. Set the discriminant to zero:
Set the discriminant to zero and solve for [tex]\( m \)[/tex]:
[tex]\[
(2m + 6)^2 - 4m^2 = 0
\][/tex]
[tex]\[
4m^2 + 24m + 36 - 4m^2 = 0
\][/tex]
[tex]\[
24m + 36 = 0
\][/tex]
[tex]\[
24m = -36
\][/tex]
[tex]\[
m = -\frac{36}{24}
\][/tex]
[tex]\[
m = -\frac{3}{2}
\][/tex]
Thus, the value of [tex]\( m \)[/tex] for which the line [tex]\( mx + y + 1 = 0 \)[/tex] is tangent to the parabola [tex]\( y^2 = -6x \)[/tex] is [tex]\(-\frac{3}{2}\)[/tex].
Therefore, the correct answer is:
[tex]\[
(b) \quad m = -\frac{3}{2}
\][/tex]
