Where does the normal line to the parabola
y = x2 − 1
at the point
(−1, 0)
intersect the parabola a second time?

To find where the normal line to the parabola \( y = x^2 - 1 \) at the point \( (-1, 0) \) intersects the parabola a second time, we can follow these steps:

### Step 1: Find the slope of the tangent line
The derivative of \( y = x^2 - 1 \) gives the slope of the tangent line to the parabola at any point. The derivative is:

\[
\frac{dy}{dx} = 2x
\]

At the point \( (-1, 0) \), the slope of the tangent line is:

\[
\text{slope of tangent} = 2(-1) = -2
\]

### Step 2: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the tangent slope. So, the slope of the normal line is:

\[
\text{slope of normal} = \frac{1}{2}
\]

### Step 3: Equation of the normal line
Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line, we can find the equation of the normal line. The point \( (-1, 0) \) is on the normal line, and the slope is \( \frac{1}{2} \). Therefore, the equation of the normal line is:

\[
y - 0 = \frac{1}{2}(x + 1)
\]

Simplifying this:

\[
y = \frac{1}{2}(x + 1)
\]
\[
y = \frac{1}{2}x + \frac{1}{2}
\]

### Step 4: Find the points where the normal line intersects the parabola
We now find where this normal line intersects the parabola \( y = x^2 - 1 \) a second time. To do this, we set the equation of the normal line equal to the equation of the parabola:

\[
x^2 - 1 = \frac{1}{2}x + \frac{1}{2}
\]

Multiply the entire equation by 2 to eliminate the fractions:

\[
2x^2 - 2 = x + 1
\]

Rearranging terms:

\[
2x^2 - x - 3 = 0
\]

### Step 5: Solve the quadratic equation
We solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 24}}{4}
\]
\[
x = \frac{1 \pm \sqrt{25}}{4}
\]
\[
x = \frac{1 \pm 5}{4}
\]

So, the two solutions for \( x \) are:

\[
x = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}, \quad x = \frac{1 - 5}{4} = \frac{-4}{4} = -1
\]

### Step 6: Find the corresponding y-coordinates
We already know that the normal line passes through \( (-1, 0) \), so the second point of intersection corresponds to \( x = \frac{3}{2} \). Plug this value into the equation of the parabola to find the y-coordinate:

\[
y = \left(\frac{3}{2}\right)^2 - 1 = \frac{9}{4} - 1 = \frac{9}{4} - \frac{4}{4} = \frac{5}{4}
\]

### Final Answer:
The normal line intersects the parabola a second time at the point \( \left(\frac{3}{2}, \frac{5}{4}\right) \).

Would you like further details or have any questions?

Here are 5 related questions to explore this topic further:
1. How do you find the tangent line to any curve at a given point?
2. What is the geometric meaning of the normal line to a curve?
3. Can you derive the quadratic formula used to solve equations like \( 2x^2 - x - 3 = 0 \)?
4. What happens when the normal line doesn't intersect the parabola again? How would you interpret this geometrically?
5. How does the concept of normals change for 3D curves?

**Tip:** The normal line is always perpendicular to the tangent line at the point of intersection.

Where does the normal line to the parabola y = x^2 - 1 at the point (-1, 0) intersect the parabola a second time?

Explore how to find the point where the normal line to the parabola y = x^2 - 1 at (-1, 0) intersects the parabola a second time. This problem covers calculus concepts like derivatives, tangent and normal lines, and algebraic methods for solving quadratic equations. Suitable for students in Grades 10-12.

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