Chapter 11: Conic Sections

Long Answer type:

Since, diameters of a circle intersect at the centre of a circle,

2x – 3y = 5 ………1

3x – 4y = 7 ………..2

Solving the above equations,

Multiplying equation 1 by 3 we get

6x – 9y = 15

Multiplying equation 2 by 2 we get

6x – 8y = 14

y = 1

y = -1

Putting y = -1, in equation 1, we get

2x – 3(-1) = 5

2x + 3 = 5

2x = 2

x = 1

Coordinates of centre = (1,-1)

Given area = 154

Area = πr² = 154

22/7 × r² = 154

r² = 154 × 7/22

r = 7 units

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

(x – 1)² + (y – (-1))² = 7²

x² – 2x +1 + (y + 1)² = 49

x² – 2x + 1 + y² + 2y + 1 – 49 = 0

x² – 2x + y² + 2y – 47 = 0

Hence the required equation of the given circle is x² – 2x + y² + 2y – 47 = 0.

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²………..1

Substituting (2, 3) & (4, 5) in the above equation, we get

(2 – h)² + (3 – k)² = r²

4 – 4h + h² + 9 + k² – 6k = r²

h² – 4h + k² – 6k + 13 = r² …………..2

(4 – h)² + (5 – k)² = r²

16 – 8h + h² + 25 + k² – 10k = r²

h² – 8h + k² – 10k + 41 = r² ………..3

Equating both the equations 2 & 3, as their RHS are equal, we get

h² – 4h + k² – 6k + 13 = h² – 8h + k² – 10k + 41

On simplifying we get

8h – 4h + 10k – 6k = 41 – 13

4h + 4k = 28

h + k = 7 ………..4

As centre lies on the given line, so it satisfies the values too,

k – 4h + 3 = 0…………5

Solving equations 3 and 4 simultaneously,

h + k = 7

-4h + k = -3

Subtracting both the equations, we get

5h = 10

h = 2

2 + k = 7

k = 5

Putting h = 2 & k = 5 in equation 2,

h² – 4h + k² – 6k + 13 = r²

2² – 4(2) + 5² – 6(5) + 13 = r²

4 – 8 + 25 – 30 + 13 = r²

r² = 4

r = 2 units

Putting the values of h = 2, k = 5 & r = 2, respectively in equation 1,

(x – h)² + (y – k)² = r²

(x – 2)² + (y – 5)² = 2²

x² – 4x + 4 + y² – 10y + 25 = 4

x² – 4x + y² – 10y + 25 =0

Hence, the required equation is x² – 4x + y² – 10y + 25 = 0.

Using Pythagoras Theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

= (3)² + (√29)² = 29 + 9

= √38

Hypotenuse = √38 units (radius)

Since, the radius bisects the chord into two equal halves,

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

(x – 3)² + (y – (-1))² = (√38)²

x² – 6x + 9 + (y + 1)² = 38

x² – 6x + y² + 2y + 1 + 9- 38 = 0

x² – 6x + y² + 2y – 28 = 0

Hence, the required equation of the circle is x² – 6x + y² + 2y – 28 = 0.

Given x² – 2x + y² – 4y – 20 = 0

x² – 2x + 1 +y² – 4y + 4 – 20 – 5 = 0

(x – 1)² + (y – 2)² = 25

(x – 1)² + (y – 2)² = 5²

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²

Centre = (1, 2)

Point of Intersection = (5, 5)

It intersects the line into 1: 1, as the radius of both the circles is 5 units.

Using Ratio Formula,

p + 1 = 10, q + 2 = 10

p = 9 & q = 8

Co-ordinates = (9, 8)

Therefore the equation is,

(x – h)² + (y – k)² = r²

(x – 9)² + (y – 8)² = 5²

x² – 18x + 81 + y² – 16y + 64 = 25

x² – 18x + y² – 16y + 145 – 25 = 0

x² – 18x + y² – 16y + 120 = 0

Hence, the required equation is x² – 18x + y² – 16y + 120 = 0.

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)² + (y – k)² = r²Centre lies on the line i.e., y = x – 1,

Co – Ordinates are (h, k) = (h, h – 1)

(x – h)² + (y – k)² = r²

(7 – h)² + (3 – (h – 1))² = 3²

49 + h² – 14h + (3 – h +1)² = 9

On rearranging we get

h² – 14h + 49 +16 +h² – 8h – 9 = 0

2h² – 22h + 56 = 0

h² – 11h + 28 = 0

h² – 4h – 7h + 28 = 0

h (h – 4) – 7 (h – 4) = 0

(h – 7) (h – 4) = 0

h = 7 or 4

Centre = (7, 6) or (4, 3)

(x – h)² + (y – k)² = r²

Equation, having centre (7, 6)

(x – 7)² + (y – 6)² = 3²

x² – 14x + 49 + y² – 12y + 36 – 9 = 0

x² – 14x + y² – 12y + 76 = 0

Equation, having centre (4, 3)

(x – 4)² + (y – 3)² = 3²

x² – 8x + 16 + y² – 6y + 9 – 9 = 0

x² – 8x + y² – 6y + 16 = 0

Hence, the required equation is x² – 14x + y² – 12y + 76 = 0 or x² – 8x + y² – 6y + 16 = 0.

(a) The distance of any point on the parabola from its focus and its directrix is same.

Given that, directrix, x = 0 and focus = (6, 0)

If a parabola has a vertical axis, the standard form of the equation of the parabola is (x – h)² = 4p (y – k), where p≠ 0.

The vertex of this parabola is at (h, k).

The focus is at (h, k + p) & the directrix is the line y = k – p.

As the focus lies on x – axis,

Therefore the equation is y² = 4ax or y² = -4ax

So, for any point P(x, y) on the parabola

Distance of point from directrix = Distance of point from focus

x² = (x – 6)² + y²

x² = x² – 12x + 36 + y²

y² – 12x + 36 = 0

Hence the required equation is y² – 12x + 36 = 0.