Exercise 3.6 - Chapter 3 - Analytical Geometry - 11th Business Maths Guide Samacheer Kalvi Solutions

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Exercise 3.6

Text Book Back Questions and Answers

Question 1.
Find the equation of the parabola whose focus is the point F(-1, -2) and the directrix is the line 4x – 3y + 2 = 0.
Solution:
F(-1, -2)
l : 4x – 3y + 2 = 0
Let P(x, y) be any point on the parabola.
FP = PM
⇒ FP² = PM²

⇒ 25(x² + y² + 2x + 4y + 5) = 16x² + 9y² – 24xy + 16x – 12y + 4
⇒ (25 – 16)x² + (25 – 9)y² + 24xy + (50 – 16)x + (100 + 12)y + 125 – 4 = 0
⇒ 9x² + 16y² + 24xy + 34x + 112y + 121 = 0

Question 2.
The parabola y² = kx passes through the point (4, -2). Find its latus rectum and focus.
Solution:
y² = kx passes through (4, -2)
(-2)² = k(4)
⇒ 4 = 4k
⇒ k = 1

Question 3.
Find the vertex, focus, axis, directrix, and the length of the latus rectum of the parabola y² – 8y – 8x + 24 = 0.
Solution:
y² – 8y – 8x + 24 = 0
⇒ y² – 8y – 4² = 8x – 24 + 4²
⇒ (y – 4)² = 8x – 8
⇒ (y – 4)² = 8(x – 1)
⇒ (y – 4)² = 4(2) (x – 1)
∴ a = 2
Y² = 4(2)X where X = x – 1 and Y = y – 4

Question 4.
Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola (a) y² = 20x, (b) x² = 8y, (c) x² = -16y
Solution:
(a) y² = 20x
y² = 4(5)x
∴ a = 5

(b) x² = 8y = 4(2)y
∴ a = 2

Question 5.
The average variable cost of the monthly output of x tonnes of a firm producing a valuable metal is 1/5 x² – 6x + 100. Show that the average variable cost curve is a parabola. Also, find the output and the average cost at the vertex of the parabola.
Solution:
Let output be x and average variable cost = y
y = 1/5 x² – 6x + 100
⇒ 5y = x² – 30x + 500
⇒ x² – 30x + 225 = 5y – 500 + 225
⇒ (x – 15)² = 5y – 275
⇒ (x – 15)² = 5(y – 55) which is of the form X² = 4(5/4)Y
∴ Y average variable cost curve is a parabola
Vertex (0, 0)
x – 15 = 0; y – 55 = 0
x = 15; y = 55
At the vertex, output is 15 tonnes and average cost is 55.

Question 6.
The profit y accumulated in thousand in x months is given by y = -x² + 10x – 15. Find the best time to end the project.
Solution:
y = -x² + 10x – 15
⇒ y = -[x² – 10x + 5² – 5² + 15]
⇒ y = -[(x – 5)² – 10]
⇒ y = 10 – (x – 5)²
⇒ (x – 5)² = -(y – 10)
This is a parabola which is open downwards.
Vertex is the maximum point.
∴ Profit is maximum when x – 5 = 0 (or) x = 5 months.
After that profit gradually reduces.
∴ The best time to end the project is after 5 months.