Conic Sections · JEE Preparation Mathematics
An ellipse has its foci at (-4, 0) and (4, 0), and its major axis has length 10. Find its equation.
Practice question from Conic Sections (JEE Preparation Mathematics), with the correct answer and a step-by-step solution.
medium · Subjective · Q-B0188F
An ellipse has its foci at (-4, 0) and (4, 0), and its major axis has length 10. Find its equation.
Answer: The semi-major axis is a = 10/2 = 5 and the focal distance is c = 4. Thus, b^2 = a^2 - c^2 = 25 - 16 = 9, so the equation of the ellipse is x^2/25 + y^2/9 = 1.
Solution
The centre is the midpoint of the foci, namely . Since and , we get ; therefore, the ellipse is .
More from Conic Sections
- The radius of the circle x^2 + y^2 - 6x + 8y + 9 = 0 is:
- The eccentricity of the ellipse x^2/25 + y^2/9 = 1 is:
- The equation of the tangent to the parabola y^2 = 16x at the point (4, 8) is:
- For the parabola y^2 = -12x, the focus and directrix are respectively:
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