# **Exercise Questions** All questions in this exercise are listed below. Click on a question to view its solution. --- # **Key Concepts** This exercise focuses on the following concepts: - Standard properties of parabolas (focus, vertex, axis of symmetry, directrix) - Latus rectum dimensions and endpoint coordinates - Deriving parabola equations from the focus-directrix definition - Parametric and trigonometric coordinate representations - Calculus approach to finding vertices (differentiation and extrema) --- # **Important Formulas** Below are the key formulas used in this exercise: Standard equations of parabolas: $$y^2 = 4ax \quad \text{(opens right)}, \quad x^2 = 4ay \quad \text{(opens up)}$$ $$(y-k)^2 = 4a(x-h) \quad \text{(vertex at }(h,k)\text{)}$$ Focus-directrix property: $$\text{Distance to focus} = \text{Distance to directrix}$$ $$\sqrt{(x-x_0)^2 + (y-y_0)^2} = \frac{|ax + by + c|}{\sqrt{a^2+b^2}}$$ Latus rectum: $$\text{Length} = 4a$$ $$\text{Endpoints: } (a, \pm 2a) \text{ for } y^2=4ax$$ Vertex via differentiation: $$\frac{dy}{dx} = 0 \quad \text{or} \quad \frac{dx}{dy} = 0 \text{ at vertex}$$ --- # **Summary** This exercise covers parabolas through both geometric and algebraic lenses. It begins with identifying key features (focus, directrix, latus rectum) from equations, then progresses to constructing equations from geometric constraints—including parametric cases with trigonometric coordinates. The final question bridges algebra and calculus, demonstrating that the vertex represents the extremum point found via differentiation. Key strategies include applying the focus-directrix distance equality and recognizing standard forms.