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1.3 – N – CONICS – Exercise Problems with solutions

Part – A

1. Prove that the equation  x²  –  2xy  + y² – 16x – 12y + 22 = 0 is a parabola. 

Soln:    x²  –  2xy  + y² – 16x – 12y + 22 = 0                    —————–   ( 1 )

Condition for  ( 1 ) to represent parabola is  h² = ab

From ( 1 )   a =  1,    b = 1

2h = -2  ⇒  h = -1

h² = ab

(-1)²  =   1 ( 1)

1 = 1

1 = 1.                 ∴  ( 1 )   represents a parabola.

2. Show that the equation  7x²  +  3xy  + 2y² – x + 2y – 1 = 0  represents an ellipse.

Soln:  Given    7x²  +  3xy  + 2y² – x + 2y – 1 = 0  –––– (1)

ax²  +  2hxy  + by² + 2gx + 2fy + c = 0    ––––– (2)

Comparing, we get

a = 7                2h = 3              b = 2

h = 3/2

h² –  ab =  (3/2)²  –  7(2)  =  9/4 – 14 =  (9 -56)/14 = -47/14 < 0

Given equation ( 1 ) represents ellipse.

Part – C

1. Find axis, vertex, focus and equation of directrix for y² + 4x + 6y + 17 = 0.

Soln:   y² + 4x + 6y + 17 = 0  

y² + 6y = -4x – 17

y² + 6y + 9 = -4x – 17 + 9   ( Adding 9 on bothsides)

(y + 3)² = – 4x – 8

= – 4(x + 2)

(y + 3)² =     – 4(x + 2)

This is of the form  Y²  =  – 4X                               (  y²  =  -4ax)  (open leftward)

Where  Y = y – 3   and   X  =  x + 2

4a  = 4

a =  1

vertex  (0 , 0) for X , Y

Y = 0  ⇒   y + 3   = 0

y = -3

X  = 0    ⇒    x + 2 = 0

x = -2

The vertex is ( – 2,-3)

Focus: 

The focus (X = – a, Y = 0)

X  = -a

x + 2  = – 1

x  = – 3

Y = 0  ⇒   y +3   = 0

y = – 3

Focus is ( -3, – 3 )

Equation of directrix is  X – a = 0

x + 2 – 1 = 0

x + 1 = 0

Latus rectum X + a = 0

x + 2 + 1 = 0

x + 3 = 0

Length of latus rectum = 4a = 4(1) = 4.