To find the slope of the third normal from point P on the parabola given by the equation \(y^2 = 8x\), we start by analyzing the properties of normals to the parabola. The normals at points on the parabola can be derived from the general equation of the normal line.
Understanding the Parabola
The parabola \(y^2 = 8x\) opens to the right. For a point \(P(x_1, y_1)\) on this parabola, the coordinates satisfy the equation, meaning \(y_1^2 = 8x_1\).
Finding the Slope of Normals
The slope of the normal to the parabola at any point \((x, y)\) is given by:
- Normal slope = \(-\frac{y}{4}\) (derived from the derivative of the parabola).
For the point \(P(x_1, y_1)\), the slope of the normal is:
m = -\frac{y_1}{4}
Conditions for Right Angles
If two normals from point P are at right angles, their slopes \(m_1\) and \(m_2\) satisfy:
m_1 \cdot m_2 = -1
Let’s denote the slopes of the two normals as:
- m₁ = -\frac{y_1}{4}
- m₂ = -\frac{y_1}{4} + k (for some k)
From the right angle condition, we can derive that:
m_1 \cdot m_2 = -1
Finding the Third Normal's Slope
The slope of the third normal can be derived from the relationship between the slopes of the normals. Since the slopes of the normals are related through the angles they form, we can conclude that:
The slope of the third normal will be:
m_3 = -\frac{y_1}{2}
Final Answer
Thus, the slope of the third normal is:
−y₁/2
The correct option is D: −y₁/2.
