Circle, Ellipse, Parabola & Hyperbola: Solved Examples

The Standard Form of Circle

• (x - h)² + (y - k)² = r²

  The centre of the circle is (h, k) and the radius is r.

• x² + y² = r²

  This is the equation when the centre is at the origin (0, 0). It is the simplest form of the circle's equation.

Example 1: If the area of the circle shown below is kπ, what is the value of k?
(a) 4   
(b) 16   
(c) 32   
(d) 20

Sol.

The radius of the circle is the distance between the points (4, 4) and (0, 0).

r² = (4 - 0)² + (4 - 0)²

r² = 16 + 16 = 32

Area = πr² = 32π

Therefore, k = 32.

The Ellipse

Ellipse

• An ellipse is the locus of a point such that the sum of its distances from two fixed points (foci) is constant.

• Standard equation (centre at origin):

  
  

  Here, a and b are the semi-axes. If a ≥ b, then the major axis lies along the x-axis and the vertices are (±a, 0) and (0, ±b).

  The foci are at (±c, 0) where c² = a² - b².

  The eccentricity is e = c/a with 0 ≤ e < 1.

The Parabola

Parabola

• A parabola is the locus of a point that moves so that its distance from a fixed point (focus) equals its distance from a fixed line (directrix). Parabolic paths appear in projectile motion (for example, a baseball hit into the air) and many engineering applications (reflectors, antennas).

• A parabola is represented by any second-degree equation in x and y where exactly one variable is squared e.g., y = ax² + bx + c or x = ay² + by + c.

  The graph of any non-degenerate quadratic in x is a parabola. It opens upwards if a > 0 and downwards if a < 0. The axis of symmetry is vertical (parallel to the y-axis) for this form.

• Standard forms:

  y² = 4ax - parabola with axis along the x-axis and vertex at origin (focus at (a, 0)).

  x² = 4ay - parabola with axis along the y-axis and vertex at origin (focus at (0, a)).

• Examples: y = x² - 2x + 1 and y = -x² - 4 are parabolic equations.

The Hyperbola

Hyperbola

• A hyperbola is formed when a right circular cone is cut by a plane parallel to the axis of the cone. It consists of two disconnected branches.

• A hyperbola is the locus of a point such that the absolute difference of its distances from two fixed points (foci) is constant.

• Standard equation (centre at origin, transverse axis on x-axis):

• 
  The foci are at (±c, 0) where c² = a² + b².
  The asymptotes are y = ±(b/a) x.
  
  
  

Example 2: Find the area enclosed by the figure | x | + | y | = 4.

Sol. 

The four possible lines are:
x + y = 4; x - y = 4; - x - y = 4 and -x + y = 4
The four lines can be represented on the coordinates axes as shown in the figure. Thus a square is formed with the vertices as shown. The side of the square is:
The area of the square is = 32 sq. units.

Example 3: If point (t, 1) lies inside circle x² + y² = 10, then t must lie between:

As (t, 1) lies inside the circle, so its distance from centre i.e. origin should be less than radius i.e.

Example 4: Find the equation of line passing through (2, 4) and through the intersection of lines 4x - 3y - 21 = 0 and 3x - y - 12 = 0?

Sol. Find the intersection point of the two given lines by solving simultaneously.

From 3x - y - 12 = 0, we get y = 3x - 12.

Substitute into 4x - 3y - 21 = 0:

4x - 3(3x - 12) - 21 = 0

4x - 9x + 36 - 21 = 0

-5x + 15 = 0

x = 3

y = 3(3) - 12 = -3

The point of intersection is (3, -3).

Now find the equation of the line through (3, -3) and (2, 4).

Slope m = (4 - (-3)) / (2 - 3)

m = 7 / (-1) = -7

Using point-slope form: y + 3 = -7(x - 3)

Simplify: y + 3 = -7x + 21

Bring all terms to one side: 7x + y - 18 = 0

Alternate Method:
Equation of line through intersection of 4x - 3y - 21 = 0 and 3x - y - 12 = 0 is:
(4x - 3y - 21) + k(3x - y - 12) = 0.
As this line passes through (2, 4):
⇒ (4 × 2 - 3 × 4 - 21) + k(3 × 2 - 4 - 12) = 0  
⇒ k =

Key Formulae and Properties

• Circle: (x - h)² + (y - k)² = r²; centre (h, k); radius r.
• Ellipse: x²/a² + y²/b² = 1; foci at (±c, 0) with c² = a² - b²; eccentricity e = c/a (0 ≤ e < 1).
• Parabola: y = ax² + bx + c is a parabola; standard forms y² = 4ax or x² = 4ay; focus-directrix property.
• Hyperbola: x²/a² - y²/b² = 1; foci at (±c, 0) with c² = a² + b²; asymptotes y = ±(b/a) x; eccentricity e = c/a (> 1).