Chapter 9: Problem 68
Derive the standard equation of a parabola opening right, \(y^{2}=4 p x .\) [Calculate the distance \(d_{1}\) from any point on the parabola \((x, y)\) to the focus \((p, 0) .\) Calculate the distance \(d_{2}\) from any point on the parabola \((x, y)\) to the directrix \((x,-p)\) Set \(\left.d_{1}=d_{2} .\right]\)
Short Answer
Step by step solution
Calculate Distance to the Focus
Calculate Distance to the Directrix
Equate the Distances
Simplify the Equation
Solve for y^2
Conclusion of Simplification
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Focus and Directrix
- Denoted by \((p, 0)\), where \(p\) is a constant that determines the distance from the vertex to the focus.
- The vertical line \(x = -p\), with the same constant \(p\) used in the focus.
Distance Formula
- \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
- \(d_1 = \sqrt{(x - p)^2 + y^2}\)
- \(d_2 = |x + p|\)
Completing the Square
- We start by expanding and manipulating the equation \((x - p)^2 + y^2 = (x + p)^2\).
- \(x^2 - 2px + p^2 + y^2 = x^2 + 2px + p^2\)
- \(-2px + y^2 = 2px\)
- \(y^2 = 4px\)