Exercise 3.7 - Chapter 3 Analytical Geometry 11th Business Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Exercise 3.7

 Text Book Back Questions and Answers

Question 1.
If m₁ and m₂ are the slopes of the pair of lines given by ax² + 2hxy + by² = 0, then the value of m₁ + m₂ is:

Question 2.
The angle between the pair of straight lines x² – 7xy + 4y² = 0 is:

Hint:
x² – 7xy + 4y² = 0
Here 2h = -7, a = 1, b = 4

Question 3.
If the lines 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0 are the diameters of a circle, then its centre is:
(a) (-1, 1)
(b) (1, 1)
(c) ( 1, -1)
(d) (-1, -1)
Answer:
(c) ( 1, -1)
Hint:
To get centre we must solve the given equations
2x – 3y – 5 = 0 …….(1)
3x – 4y – 7 = 0 ………(2)
(1) × 3 ⇒ 6x – 9y = 15
(2) × 2 ⇒ 6x – 8y = 14
Subtracting, -y = 1 ⇒ y = -1
Using y = -1 in (1) we get
2x + 3 – 5 = 0
⇒ 2x = 2
⇒ x = 1

Question 4.
The x-intercept of the straight line 3x + 2y – 1 = 0 is
(a) 3
(b) 2

Hint:
To get x-intercept put y = 0 in 3x + 2y – 1 = 0 we get
3x – 1 = 0
x = 1/3

Question 5.
The slope of the line 7x + 5y – 8 = 0 is:

Question 6.
The locus of the point P which moves such that P is at equidistance from their coordinate axes is:

Answer:
(c) y = x
Hint:

Given PA = PB
y₁ = x₁
∴ Locus is y = x

Question 7.
The locus of the point P which moves such that P is always at equidistance from the line x + 2y + 7 = 0:
(a) x + 2y + 2 = 0
(b) x – 2y + 1 = 0
(c) 2x – y + 2 = 0
(d) 3x + y + 1 = 0
Answer:
(a) x + 2y + 2 = 0
Hint:
Locus is line parallel to line x + 2y + 7 = 0 which is x + 2y + 2 = 0

Question 8.
If kx² + 3xy – 2y² = 0 represent a pair of lines which are perpendicular then k is equal to:

(c) 2
(d) -2
Answer:
(c) 2
Hint:
Here a = k, b = -2
Condition for perpendicular is
a + b = 0
⇒ k – 2 = 0
⇒ k = 2

Question 9.
(1, -2) is the centre of the circle x² + y² + ax + by – 4 = 0, then its radius:
(a) 3
(b) 2
(c) 4
(d) 1
Answer:
(a) 3
Hint:
Given centre (-g, -f) = (1, -2)
From the given equation c = -4

Question 10.
The length of the tangent from (4, 5) to the circle x² + y² = 16 is:
(a) 4
(b) 5
(c) 16
(d) 25
Answer:
(b) 5
Hint:
Length of the tangent from (x₁, y₁) to the circle x² + y² = 16 is 

Question 11.
The focus of the parabola x² = 16y is:
(a) (4 , 0)
(b) (-4, 0)
(c) (0, 4)
(d) (0, -4)
Answer:
(c) (0, 4)
Hint:
x² = 16y
Here 4a = 16 ⇒ a = 4
Focus is (0, a) = (0, 4)

Question 12.
Length of the latus rectum of the parabola y² = -25x:
(a) 25
(b) -5
(c) 5
(d) -25
Answer:
(a) 25
Hint:
y² = -25a
Here 4a = 25 which is the length of the latus rectum.

Question 13.
The centre of the circle x² + y² – 2x + 2y – 9 = 0 is:
(a) (1, 1)
(b) (-1, 1)
(c) (-1, 1)
(d) (1, -1)
Answer:
(d) (1, -1)
Hint:
2g = -2, 2f = 2
g = -1, f = 1
Centre = (-g, -f) = (1, -1)

Question 14.
The equation of the circle with centre on the x axis and passing through the origin is:
(a) x² – 2ax + y² = 0
(b) y² – 2ay + x² = 0
(c) x² + y² = a²
(d) x² – 2ay + y² = 0
Answer:
(a) x² – 2ax + y² = 0
Hint:
Let the centre on the x-axis as (a, 0).
This circle passing through the origin so the radius

Now centre (h, k) = (a, 0)
Radius = a
Equation of the circle is (x – a)² + (y – 0)² = a²
⇒ x² – 2ax + a² + y² = a²
⇒ x² – 2ax + y² = 0

Question 15.

If the centre of the circle is (-a, -b) and radius is  then the equation of circle is:

(a) x² + y² + 2ax + 2by + 2b² = 0
(b) x² + y² + 2ax + 2by – 2b² = 0
(c) x² + y² – 2ax – 2by – 2b² = 0
(d) x² + y² – 2ax – 2by + 2b² = 0
Answer:
(a) x² + y² + 2ax + 2by + 2b² = 0
Hint:
Equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x + a)² + (y + b)² = a² – b²
⇒ x² + y² + 2ax + 2by + a² + b² = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

Question 16.
Combined equation of co-ordinate axes is:
(a) x² – y² = 0
(b) x² + y² = 0
(c) xy = c
(d) xy = 0
Answer:
(d) xy = 0
Hint:
Equation of x-axis is y = 0
Equation of y-axis is x = 0
Combine equation is xy = 0

Question 17.
ax² + 4xy + 2y² = 0 represents a pair of parallel lines then ‘a’ is:
(a) 2
(b) -2
(c) 4
(d) -4
Answer:
(a) 2
Hint:
Here a = 0, h = 2, b = 2
Condition for pair of parallel lines is b² – ab = 0
4 – a(2) = 0
⇒ -2a = -4
⇒ a = 2

Question 18.
In the equation of the circle x² + y² = 16 then v intercept is (are):
(a) 4
(b) 16
(c) ±4
(d) ±16
Answer:
(c) ±4
Hint:
To get y-intercept put x = 0 in the circle equation we get
0 + y² = 16
∴ y = ±4

Question 19.
If the perimeter of the circle is 8π units and centre is (2, 2) then the equation of the circle is:
(a) (x – 2)² + (y – 2)² = 4
(b) (x – 2)² + (y – 2)² = 16
(c) (x – 4)² + (y – 4)² = 16
(d) x² + y² = 4
Answer:
(c) (x – 2)² + (y – 2)² = 16
Hint:
Perimeter, 2πr = 8π
r = 4
Centre is (2, 2)
Equation of the circle is (x – 2)² + (y – 2)² = 4² = 16

Question 20.
The equation of the circle with centre (3, -4) and touches the x-axis is:
(a) (x – 3)² + (y – 4)² = 4
(b) (x – 3)² + (y + 4)² = 16
(c) (x – 3)² + (y – 4)² = 16
(d) x² + y² = 16
Answer:
(b) (x – 3)² + (y + 4)² = 16
Hint:
Centre (3, -4).
It touches the x-axis.
The absolute value of y-coordinate is the radius, i.e., radius = 4.
Equation is (x – 3)² + (y + 4)² = 16

Question 21.
If the circle touches the x-axis, y-axis, and the line x = 6 then the length of the diameter of the circle is:
(a) 6
(b) 3
(c) 12
(d) 4
Answer:
(a) 6
Hint:

Question 22.
The eccentricity of the parabola is:
(a) 3
(b) 2
(c) 0
(d) 1
Answer:
(d) 1

Question 23.
The double ordinate passing through the focus is:
(a) focal chord
(b) latus rectum
(c) directrix
(d) axis
Answer:
(b) latus rectum
Hint:

Question 24.
The distance between directrix and focus of a parabola y² = 4ax is:
(a) a
(b) 2a
(c) 4a
(d) 3a
Answer:
(b) 2a

Question 25.
The equation of directrix of the parabola y² = -x is:
(a) 4x + 1 = 0
(b) 4x – 1 = 0
(c) x – 1 = 0
(d) x + 4 = 0
Answer:
(b) 4x – 1 = 0
Hint:
y² = -x.
It is a parabola open leftwards.