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369 LearnersLast updated on August 5, 2025

Conic sections include circles, ellipses, parabolas, and hyperbolas. Let’s explore the formulas used to describe these conic sections.


A parabola is the set of all points in the plane that are equidistant from a fixed point (focus) and a given line (directrix). The standard equation of a parabola with vertex at (h, k) is:
[y - k = a(x - h)^2] (for a vertical parabola) [x - h = a(y - k)^2] (for a horizontal parabola)
A hyperbola is the set of all points where the difference of the distances from two fixed points (foci) is constant. The standard equation of a hyperbola centered at (h, k) is:
[frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1] for a horizontal hyperbola, and [frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} = 1] for a vertical hyperbola.
Conic sections are fundamental in mathematics and have numerous real-life applications.
These formulas help describe planetary orbits, design optical lenses, and analyze satellite paths.
By learning these formulas, students can understand key concepts in geometry, physics, and engineering.
Students make errors when working with conic sections. Here are some mistakes and ways to avoid them.
Find the equation of a circle with center (3, -2) and radius 5.
The equation is \((x - 3)^2 + (y + 2)^2 = 25\)
Using the formula: \((x - h)^2 + (y - k)^2 = r^2\) Center (h, k) = (3, -2) and radius r = 5. Thus, \((x - 3)^2 + (y + 2)^2 = 25\).
Write the standard form of an ellipse with center (0, 0), a = 4, b = 2.
The equation is \(\frac{x^2}{16} + \frac{y^2}{4} = 1\)
For an ellipse centered at the origin with a > b: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) Here, a = 4, b = 2, so: \(\frac{x^2}{16} + \frac{y^2}{4} = 1\).
Find the equation of a parabola with vertex (0, 0) and focus at (0, 3).
The equation is \(y = \frac{1}{12}x^2\)
Since the parabola opens upwards, use: \[y = \frac{1}{4p}x^2\] Here, p = 3, so: \[y = \frac{1}{12}x^2\].
Determine the equation of a hyperbola with center (0, 0), a = 3, b = 4.
The equation is \(\frac{x^2}{9} - \frac{y^2}{16} = 1\)
For a hyperbola centered at the origin with a horizontal transverse axis: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) Here, a = 3, b = 4, so: \(\frac{x^2}{9} - \frac{y^2}{16} = 1\).
Find the equation of a circle with radius 7 and center at (5, -4).
The equation is \((x - 5)^2 + (y + 4)^2 = 49\)
Using the formula: \((x - h)^2 + (y - k)^2 = r^2\) Center (h, k) = (5, -4) and radius r = 7. Thus, \((x - 5)^2 + (y + 4)^2 = 49\).
Vikrant is a passionate Mathematics teacher with over 5 years of teaching experience in both conventional and Vedic Maths. His student-focused approach combines clear concepts with motivation and encouragement, helping students build confidence and develo
: He loves to play the quiz with kids through algebra to make kids love it.
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