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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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The circle is the locus of a point that moves such that its distance from a fixed point (center) is constant (radius). In standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, the center is (h,k)(h, k). If the center is at the origin, it simplifies to x2+y2=r2x^2 + y^2 = r^2.

A circle centered at the origin with radius r.
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A parabola is the set of all points equidistant from a fixed point (focus) and a fixed line (directrix). For y2=4axy^2 = 4ax, the focus is (a,0)(a, 0) and the directrix is x=−ax = -a.

Parabola opening rightwards with focus and directrix shown.
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An ellipse is defined by ax2+by2=1ax^2 + by^2 = 1. The eccentricity e<1e < 1 determines how 'flat' the ellipse is. The sum of distances from any point on the ellipse to the two foci is constant and equal to the major axis length 2a2a.

Horizontal ellipse showing the positions of the two foci.
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A hyperbola x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 has two branches. The distance between the vertices is 2a2a, and the eccentricity e>1e > 1.

Hyperbola with two branches opening left and right.

📐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.

Problem 3:

Find the equation of the parabola with vertex at (0,0)(0,0) and focus at (0,3)(0, 3). Also find the equation of its directrix.

Parabola opening upwards with focus at (0,3) and directrix at y=-3.

Solution:

  1. Since the vertex is (0,0)(0,0) and the focus (0,3)(0,3) lies on the y-axis, the parabola opens upwards.
  2. The standard form is x2=4ayx^2 = 4ay.
  3. Here, a=3a = 3 (distance from vertex to focus).
  4. Equation: x2=4(3)y⇒x2=12yx^2 = 4(3)y \Rightarrow x^2 = 12y.
  5. The directrix is a horizontal line at distance aa below the vertex: y=−a⇒y=−3y = -a \Rightarrow y = -3.

Explanation:

Because the focus has a non-zero y-coordinate and zero x-coordinate, the axis of symmetry is the y-axis. The value of 'a' is the y-coordinate of the focus.

Problem 4:

Find the eccentricity, length of the latus rectum, and foci of the hyperbola 9x2−16y2=1449x^2 - 16y^2 = 144.

Hyperbola with branches opening horizontally and foci marked at 5 and -5.

Solution:

  1. Divide by 144 to get standard form: x216−y29=1\frac{x^2}{16} - \frac{y^2}{9} = 1.
  2. Here a2=16⇒a=4a^2 = 16 \Rightarrow a = 4 and b2=9⇒b=3b^2 = 9 \Rightarrow b = 3.
  3. Eccentricity e=1+b2a2=1+916=2516=54e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}.
  4. Foci are (±ae,0)=(±4×54,0)=(±5,0)(\pm ae, 0) = (\pm 4 \times \frac{5}{4}, 0) = (\pm 5, 0).
  5. Length of Latus Rectum =2b2a=2(9)4=4.5= \frac{2b^2}{a} = \frac{2(9)}{4} = 4.5.

Explanation:

First convert the equation to the form x2/a2−y2/b2=1x^2/a^2 - y^2/b^2 = 1 to identify a and b. Then use the standard hyperbola formulas for eccentricity and focal coordinates.