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Coordinate Geometry - Conic Sections (Circle, Parabola, Ellipse, Hyperbola)

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A conic section is defined as the locus of a point that moves such that the ratio of its distance from a fixed point (Focus, SS) to its distance from a fixed straight line (Directrix, LL) is a constant called eccentricity (ee). Visually, this creates different shapes depending on the value of ee: a circle (e=0e=0), a parabola (e=1e=1), an ellipse (0<e<10 < e < 1), and a hyperbola (e>1e > 1).

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The Circle: A circle is a conic where the plane intersects the cone perpendicular to its axis. It is the locus of all points at a fixed distance rr from a fixed center (h,k)(h, k). Visually, it is perfectly symmetrical, with every point on the circumference being equidistant from the center.

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The Parabola: A parabola is formed when the plane is parallel to the slant height of the cone. It consists of a single open curve. Key visual elements include the Vertex (turning point), the Focus (inside the curve), and the Directrix (a line behind the curve). For y2=4axy^2 = 4ax, the curve opens towards the positive xx-axis and is symmetric about the xx-axis.

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The Ellipse: An ellipse is a closed loop formed when the plane cuts through the cone at an angle. It has two focal points (SS and S′S') and two axes of symmetry: the Major Axis (the longer diameter) and the Minor Axis (the shorter diameter). The sum of the distances from any point on the ellipse to the two foci is always constant (2a2a).

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The Hyperbola: A hyperbola consists of two separate, mirrored curves called branches, formed when the plane cuts both nappes of a double cone. It features a Transverse Axis (connecting the vertices) and a Conjugate Axis. The difference between the distances from any point on the hyperbola to the two foci is a constant (2a2a).

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Latus Rectum: For all conics, the latus rectum is a chord passing through the focus and perpendicular to the principal axis. Visually, it measures the 'width' of the conic at the focus. Its length varies: 4a4a for a parabola, 2b2a\frac{2b^2}{a} for an ellipse or hyperbola.

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General Second-Degree Equation: The equation ax2+2hxy+by2+2gx+2fy+c=0ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 represents a conic section. The type of conic is determined by the discriminant h2−abh^2 - ab. If h2−ab=0h^2 - ab = 0, it is a parabola; if h2−ab<0h^2 - ab < 0, it is an ellipse; if h2−ab>0h^2 - ab > 0, it is a hyperbola.

📐Formulae

Circle (Standard Form): (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.

Circle (General Form): x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, Center =(−g,−f)= (-g, -f), Radius =g2+f2−c= \sqrt{g^2 + f^2 - c}.

Parabola (Standard Form): y2=4axy^2 = 4ax, Focus =(a,0)= (a, 0), Directrix: x=−ax = -a, Length of Latus Rectum =4a= 4a.

Ellipse (Standard Form): x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a>ba > b), Eccentricity e=1−b2a2e = \sqrt{1 - \frac{b^2}{a^2}}, Foci =(±ae,0)= (\pm ae, 0).

Hyperbola (Standard Form): x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, Eccentricity e=1+b2a2e = \sqrt{1 + \frac{b^2}{a^2}}, Foci =(±ae,0)= (\pm ae, 0).

Length of Latus Rectum (Ellipse/Hyperbola): LR=2b2aLR = \frac{2b^2}{a}.

Condition for Tangency to Circle x2+y2=r2x^2 + y^2 = r^2: The line y=mx+cy = mx + c is tangent if c2=r2(1+m2)c^2 = r^2(1 + m^2).

💡Examples

Problem 1:

Find the center and radius of the circle represented by the equation x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0.

Solution:

  1. Compare the given equation with the general form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0.
  2. Identify coefficients: 2g=−6  ⟹  g=−32g = -6 \implies g = -3; 2f=4  ⟹  f=22f = 4 \implies f = 2; c=−12c = -12.
  3. Calculate Center (−g,−f)(-g, -f): Center =(3,−2)= (3, -2).
  4. Calculate Radius r=g2+f2−cr = \sqrt{g^2 + f^2 - c}: r=(−3)2+(2)2−(−12)=9+4+12=25=5r = \sqrt{(-3)^2 + (2)^2 - (-12)} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5.

Explanation:

To find the circle's properties, we identify the parameters g,f,cg, f, c from the general equation and apply the standard formulas for center and radius.

Problem 2:

Find the eccentricity and the coordinates of the foci for the ellipse x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1.

Solution:

  1. Identify a2a^2 and b2b^2: a2=25  ⟹  a=5a^2 = 25 \implies a = 5; b2=16  ⟹  b=4b^2 = 16 \implies b = 4.
  2. Since a>ba > b, the major axis is along the xx-axis.
  3. Calculate eccentricity ee: e=1−b2a2=1−1625=925=35=0.6e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} = 0.6.
  4. Calculate Foci (±ae,0)(\pm ae, 0): ae=5×35=3ae = 5 \times \frac{3}{5} = 3. Foci =(±3,0)= (\pm 3, 0).

Explanation:

For an ellipse, we first determine the major axis by comparing aa and bb. Then we use the eccentricity formula for a>ba > b and find the focus distance aeae from the center.