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Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC

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The Standard Form of Circle
Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC

  • (x - h)2 + (y - k)2 = r2
    Where the center is (h, k) and radius is ‘r’.
  • If center of the circle is at the origin and the radius is 'r', then the equation of the circle is:
    x2 + y2 = r2
  • This is also known as the simplest form.

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

Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC

Correct Answer is Option (c)

Since the line segment joining (4, 4) and (0, 0) is a radius of the circle:
r2 = 42 + 42 = 32
Therefore, area = πr2 = 32π ⇒ k = 32 


Question for Solved Examples: Circle, Ellipse, Parabola & Hyperbola
Try yourself:What is the coordinates of the centre and the radius of the circle with equation: 
(x - 4)2 + (y - 3)2 = 25 
View Solution

The Ellipse

  • Though not so simple as the circle, the ellipse is nevertheless the curve most often "seen" in everyday life. The reason is that every circle, viewed obliquely, appears elliptical.
    Ellipse
    Ellipse
  • The equation of an ellipse centered at the origin and with an axial intersection at (±, a, 0) and (± b, 0) is:
    Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC

The Parabola

  • When a baseball is hit into the air, it follows a parabolic path. There are all kinds of parabolas, and there’s no simple, general parabola formula for you to memorize.
    Parabola
    Parabola
  • You should know, however, that the graph of the general quadratic equation:
    y = ax2 + bx + c is a parabola. 
  • It’s one that opens up either on top or on the bottom, with an axis of symmetry parallel to the y-axis. The graph of the general quadratic equation y = ax2 + bx + c is a parabola.
    Examples: y = x2 - 2x + 1 and y = - x2 - 4 are examples of some parabolic equations.

The Hyperbola

  • If a right circular cone is intersected by a plane parallel to its axis, part of a hyperbola is formed.
    Hyperbola
    Hyperbola
  • The equation of a hyperbola at the origin and with foci on the x-axis is:
    Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC


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

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: Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC
The area of the square is Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC= 32 sq. units.

Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC


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.Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC
Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC


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

4x - 3y - 21 = 0 …..(1)
3x - y - 12 = 0 ….(2)
Solving (1) and (2), we get point of intersection as x = 3 & y = - 3.
Now we have two points (3, -3) & (2, 4)
⇒ Slope of line m = Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC = - 7
So, the equation of line is:
⇒ y + 3 = - 7 (x - 3)        
⇒ 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 = Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC
So, the equation of line is:

Solved Examples: Circle, Ellipse, Parabola & Hyperbola - Notes | Study CSAT Preparation for UPSC CSE - UPSC

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