To solve this problem, let's break it down step by step:

Problem Analysis:

• Given Data:

  • The line $L$ passes through the points $(0, 0)$ and $(1, 2)$.
  • A line parallel to $L$ intersects the curve $y = 2x^2 + 3x + 1$ at only one point $P$.

• Objective: Determine the point $P$ where this happens.

Step 1: Find the Slope of Line $L$

The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Substituting the points $(0, 0)$ and $(1, 2)$: $m = \frac{2 - 0}{1 - 0} = 2$ So, the slope of $L$ is $m = 2$.

Step 2: Equation of the Parallel Line

A line parallel to $L$ will also have a slope of $m = 2$. The equation of a line with slope $m$ is: $y = mx + c$ Substituting $m = 2$, the equation becomes: $y = 2x + c$ Here, $c$ is the $y$-intercept of the parallel line.

Step 3: Intersection with the Curve

The curve is given as: $y = 2x^2 + 3x + 1$ At the point of intersection with the parallel line, the $y$-values must be equal: $2x + c = 2x^2 + 3x + 1$ Rearranging this equation: $2x^2 + x + (1 - c) = 0$

Step 4: Condition for a Single Point of Intersection

For the line to meet the curve at only one point, the quadratic equation must have exactly one solution. This happens when the discriminant ($\Delta$) of the quadratic equation is zero. The discriminant of $ax^2 + bx + c = 0$ is: $\Delta = b^2 - 4ac$ Here, $a = 2$, $b = 1$, and $c = 1 - c$. Substituting these values: $\Delta = (1)^2 - 4(2)(1 - c)$ $\Delta = 1 - 8 + 8c$ $\Delta = -7 + 8c$ For a single point of intersection, set $\Delta = 0$: $-7 + 8c = 0$ $8c = 7 \quad \implies \quad c = \frac{7}{8}$

Step 5: Find the Point of Intersection

The equation of the parallel line is: $y = 2x + \frac{7}{8}$ Substitute this into the curve equation $y = 2x^2 + 3x + 1$: $2x + \frac{7}{8} = 2x^2 + 3x + 1$ Rearranging terms: $2x^2 + x + \frac{1}{8} = 0$ Multiply through by 8 to eliminate the fraction: $16x^2 + 8x + 1 = 0$ This is a perfect square trinomial: $(4x + 1)^2 = 0$ So: $4x + 1 = 0 \quad \implies \quad x = -\frac{1}{4}$ Substitute $x = -\frac{1}{4}$ into the line equation $y = 2x + \frac{7}{8}$: $y = 2\left(-\frac{1}{4}\right) + \frac{7}{8}$ $y = -\frac{1}{2} + \frac{7}{8} = \frac{4}{8} + \frac{7}{8} = \frac{3}{8}$

Final Answer:

The point of intersection is: $P = \left(-\frac{1}{4}, \frac{3}{8}\right)$ Thus, the correct option is B.

Would you like more details or have any questions?

Related Questions:

1. What is the geometric significance of having the discriminant equal to zero in such problems?
2. How can you verify that the quadratic equation is a perfect square trinomial?
3. How do you derive the general equation of a line parallel to a given line?
4. What is the effect of changing the slope on the number of intersection points with the curve?
5. Can you derive the condition for a tangent line to a parabola using calculus?

Tip:

Always use the discriminant condition ($\Delta = 0$) for identifying tangency or single-point intersection cases in quadratic equations.