Which of the following gives the equations of both lines through the point $\left(2,,,,-,3\right)$ that are tangent to the parabola $y=x^2+x$?

• A. $y=2x-7$ and $y=-x+1$
• B. $y=5x-13$ and $y=-3x+3$
• C. $y=4x-11$ and $y=-2x+1$
• D. $y=x^2+x$ and $y=-x^2-x$

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the problem. We need to find the equations of two lines that pass through the point (2, -3) and are tangent to the parabola y = x^2 + x. Tangent lines touch the curve at exactly one point, and their slopes match the derivative of the curve at that point.
2. Step 2: Compute the derivative of the parabola y = x^2 + x to find the slope of the tangent lines. The derivative is given by \( \frac{dy}{dx} = 2x + 1 \). This represents the slope of the parabola at any point \( x \).
3. Step 3: Set up the equation of a line passing through the point (2, -3) with slope \( m \). The point-slope form of the equation is \( y + 3 = m(x - 2) \), which simplifies to \( y = m(x - 2) - 3 \).
4. Step 4: To find the tangent lines, the line must intersect the parabola at exactly one point. Substitute \( y = m(x - 2) - 3 \) into \( y = x^2 + x \) to find the intersection points. This gives \( m(x - 2) - 3 = x^2 + x \). Rearrange into a standard quadratic form: \( x^2 + (1 - m)x + (2m - 3) = 0 \).
5. Step 5: For the line to be tangent, the quadratic equation \( x^2 + (1 - m)x + (2m - 3) = 0 \) must have exactly one solution. This happens when the discriminant \( \Delta \) of the quadratic equation is zero. The discriminant is given by \( \Delta = b^2 - 4ac \), where \( a = 1 \), \( b = 1 - m \), and \( c = 2m - 3 \). Solve \( \Delta = 0 \) to find the values of \( m \), and use these slopes to write the equations of the tangent lines.