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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 6 Applications of Vector Algebra Ex 6.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.2

Question 1.
If \(\overline { a }\) = \(\hat { i }\) – 2\(\hat { j }\) + 3\(\hat { k }\), b = 2\(\hat { i }\) + \(\hat { j }\) – 2\(\hat { k }\), c = 3\(\hat { i }\) + 2\(\hat { j }\) + \(\hat { k }\) find \(\overline { a }\).(\(\overline { b}\) × \(\overline { c }\)).
Solution:
\(\overline { a }\).(\(\overline { b}\) × \(\overline { c }\)) = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = \(\left|\begin{array}{ccc}
1 & -2 & 3 \\
2 & 1 & -2 \\
3 & 2 & 1
\end{array}\right|\)
= 1(1 + 4) + 2(2 + 6) + 3(4 – 3)
= 5 + 16 + 3 = 24

Question 2.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors -6\(\hat { i }\) + 14\(\hat { j }\) + 10\(\hat { k }\), 14\(\hat { i }\) – 10\(\hat { j }\) – 6\(\hat { k }\) and 2\(\hat { i }\) + 4\(\hat { j }\) – 2\(\hat { k }\)
Solution:
Volume of the parallelepiped = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ]
= \(\left|\begin{array}{ccc}
-6 & 14 & 10 \\
14 & -10 & -6 \\
2 & 4 & -2
\end{array}\right|\)
= -6(20 + 24) -14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760
= 720 cu. units.

Question 3.
The volume of the parallelepiped whose coterminous edges are 7\(\hat { i }\) + λ\(\hat { j }\) – 3\(\hat { k }\), \(\hat { i }\) + 2\(\hat { j }\) – \(\hat { k }\), -3\(\hat { i }\) + 7\(\hat { j }\) + 5\(\hat { k }\) is 90 cubic units. Find the value of λ
Solution:
volume of the parallelepiped = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ]
\(\left|\begin{array}{ccc}
7 & \lambda & -3 \\
1 & 2 & -1 \\
-3 & 7 & 5
\end{array}\right|\) = 90
7(10 + 7) – λ(5 – 3) – 3(7 + 6) = 90
7(17) – λ(2) – 3(13) = 90
119 – 2λ – 39 = 90
2λ = 119 – 39 – 90
2λ = -10
λ = -5

Question 4.
If \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of
(\(\overline { a }\) + \(\overline { b }\)).(\(\overline { b }\) × \(\overline { c }\)) + (\(\overline { b }\) + \(\overline { c }\)).(\(\overline { c }\) × \(\overline { a }\)) + (\(\overline { c }\) + \(\overline { a }\)).(\(\overline { a }\) × \(\overline { b }\)).
Solution:
Given [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = ±4.
{(\(\overline { a }\) + \(\overline { b }\)).(\(\overline { b }\) × \(\overline { c }\)) + (\(\overline { b }\) + \(\overline { c }\)).(\(\overline { c }\) × \(\overline { a }\)) + (\(\overline { c }\) + \(\overline { a }\)).(\(\overline { a }\) × \(\overline { b }\)).}
= \(\overline { a }\)(\(\overline { b }\) × \(\overline { c }\)) + \(\overline { b }\) – (\(\overline { b }\) × \(\overline { c }\)) + \(\overline { b }\) – (\(\overline { c }\) × \(\overline { a }\)) + \(\overline { c }\) – (\(\overline { c }\) × \(\overline { a }\)) + \(\overline { c }\)(\(\overline { a }\) × \(\overline { b }\)) + \(\overline { a }\) – (\(\overline { a }\) × \(\overline { b }\))
= [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] + [ \(\overline { b }\), \(\overline { b }\), \(\overline { c }\) ] + [ \(\overline { p }\), \(\overline { c }\), \(\overline { a }\) ] + [ \(\overline { c }\), \(\overline { c }\), \(\overline { a }\) ] + [ \(\overline { c }\), \(\overline { a }\), \(\overline { b }\) ] + [ \(\overline { a }\), \(\overline { a }\), \(\overline { b }\) ]
= ± 4 + 0 ± 4 + 0 ± 4 + 0 = ± 12

Question 5.
Find the altitude of a parallelepiped determined by the vectors \(\overline { a }\) = -2\(\hat { i }\) + 5\(\hat { j }\) + 3\(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) + 3\(\hat { j }\) – 2\(\hat { k }\) and \(\overline { c }\) = -3\(\hat { i }\) + \(\hat { j }\) + 4\(\hat { k }\) if the base is taken as the parallelogram determined by \(\overline { b }\) and \(\overline { c }\).
Solution:
V = \(\overline { a }\) -(\(\overline { b }\) × \(\overline { c }\)) = [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ]
= \(\left|\begin{array}{ccc}
-2 & 5 & 3 \\
1 & 3 & -2 \\
-3 & 1 & 4
\end{array}\right|\)
= -2(12 + 2) -5(4 – 6) + 3(1 + 9)
= -2(14) -5(-2) + 3(10)
= -28 + 10 + 30 = 12
Area = |\(\overline { b }\) × \(\overline { c }\)| = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 3 & -2 \\
-3 & 1 & 4
\end{array}\right|\)
= \(\hat { i }\)(12 + 2) – \(\hat { j }\)(4 – 6) + \(\hat { k }\)(1 + 9)
= 14\(\hat { i }\) + 2\(\hat { j }\) + 10\(\hat { k }\)
|\(\overline { b }\) × \(\overline { c }\)| = \(\sqrt { 196+4+100 }\) = \(\sqrt { 300 }\)
= 10√3
Altitude h = \(\frac { V }{ Area }\) = \(\frac { 12 }{ 10√3 }\) = \(\frac { 12×√3 }{ 10×3 }\) = \(\frac { 2√3 }{ 5 }\)

Question 6.
Determine whether the three vectors 2\(\hat { i }\) + 3\(\hat { j }\) + \(\hat { k }\), \(\hat { i }\) – 2\(\hat { j }\) + 2\(\hat { k }\) and 3\(\hat { i }\) + \(\hat { j }\) + 3\(\hat { k }\) are coplanar.
Solution:
If vectors are coplanar, [ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = 0
[ \(\overline { a }\), \(\overline { b }\), \(\overline { c }\) ] = \(\left|\begin{array}{ccc}
2 & 3 & 1 \\
1 & -2 & 2 \\
3 & 1 & 3
\end{array}\right|\)
= 2(-6 – 2) -3(3 – 6) + 1(1 + 6)
= -2(-8) -3(-3) + 1(7) = -16 + 9 + 7 = 0
∴ The given vectors are coplanar

Question 7.
Let \(\overline { a }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { k }\), \(\overline { b }\) = \(\hat { i }\) and \(\overline { c }\) = c₁\(\hat { i }\) + c₂\(\hat { j }\) + c₃\(\hat { k }\). If c₁ = 1 and c₂ = 2, find c₃ such that \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) are coplanar.
Solution:
If \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) are coplanar [ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = 0
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 0 \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0
1(0) – 1(c₃) + 1(c₂) = 0
-c₃ + c₂ = 0
c₃ = c₂ = 2

Question 8.
If \(\overline { a }\) = \(\hat { i }\) – \(\hat { k }\), \(\overline { b }\) = x\(\hat { i }\) + \(\hat { j }\) + (1 – x)\(\hat { k }\) c = y\(\hat { i }\) + x\(\hat { j }\) + (1 + x – y) \(\hat { k }\) Show that [ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = 0
Solution:
[ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = \(\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|\)
= 1(1 + x – y – x + x²)-1(x² – y)
(1 + x – y – x + x² – x² + y)
= 1
There is no x and y terms
∴ [ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] depends on neither x nor y.

Question 9.
If the vectors a\(\hat { i }\) + a\(\hat { j }\) + c\(\hat { k }\), \(\hat { i }\) + \(\hat { k }\) and c\(\hat { i }\) + c\(\hat { j }\) + b\(\hat { k }\) are coplanar, prove that c is the geometric mean of a and b.
Solution:
[ \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) ] = 0
\(\left|\begin{array}{lll}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{array}\right|\) = 0
a(0 – c) – a(b – c) + c(c) = 0
-ac – ab + ac + c² = 0
c² – ab = 0
c² = ab
⇒ c in the geometric mean of a and b.

Question 10.
Let \(\overline { a }\), \(\overline { b }\) and \(\overline { c }\) be three non-zero vectors such that \(\overline { c }\) is a unit vector perpendicular to both \(\overline { a }\) and \(\overline { b }\). If the angle between \(\overline { a }\) and \(\overline { b }\) is \(\frac { π }{ 6 }\) show that [\(\overline { a }\), \(\overline { b }\) and \(\overline { c }\)]² = \(\frac { 1 }{ 4 }\) |\(\overline { a }\)|² |\(\overline { b }\)|²
Solution:

Hence proved

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.1

Question 1.
Obtain the equation of the circles with a radius of 5 cm and touching the x-axis at the origin in a general form.
Solution:

Given radius = 5 cm and the circle is touching x axis
So centre will be (0, ± 5) and radius = 5
The equation of the circle with centre (0, ± 5) and radius 5 units is
(x – 0)² + (y ± 5)² = 5²
(i.e) x² + y² ± 10 y + 25 – 25 = 0
(i.e) x² + y² ± 10y = 0

Question 2.
Find the equation of the circle with centre (2, – 1) and passing through the point (3, 6) in standard form.
Solution:
Centre = (2, -1) = (h, k)
Passing through the point (3, 6)
Equation of the circle (x – h)² + (y – k)² = r² ………. (1)
(3 – 2)² + (6 + 1)² = r²
1² + 7² = r²
1 + 49 = r²
r² = 50
(1) ⇒ (x – 2)² + (y + 1)² = 50

Question 3.
Find the equation of circles that touch both the axes and pass-through (-4, -2) in a general form.
Solution:

Since the circle touches both the axes, its centre will be (r, r) and the radius will be r.
Here centre = C = (r, r) and point on the circle is A = (-4, -2)
CA = r ⇒ CA² = r²
(i.e) (r + 4)² + (r + 2)² = r²
⇒ r² + 8r +16 + r² + 4r + 4 – r² = 0
(i.e) r² + 12r + 20 = 0
(r + 2) (r + 10) = 0
⇒ r = -2 or -10
When r = -2, the equation of the circle will be (x + 2)² + (y + 2)² = 2²
(i.e) x² + y² + 4x + 4y + 4 = 0
When r = -10, the equation of the circle will be (x + 10)² + (y + 10)² = 10²
(i.e) x² + y² + 20x + 20y + 100 = 0

Question 4.
Find the equation of the circles with centre (2, 3) and passing through the intersection of the lines 3x – 2y – 1 =0 and 4x + y – 27 = 0.
Solution:
centre (2, 3) = (h, k)
Point of intersection
Solve 3x – 2y – 1 = 0 ………. (1)
4x + y – 27 = 0 ……… (2)
(1) ⇒ 3x – 2y = 1
(2) × 2 ⇒ 8x + 2y = 54
11x = 55
x = 5
put in (1)
15 – 2y – 1 = 0
14 = 2y
y = 7
Passing-through point is (5, 7)
Equation of circle be (x – h)² + (y – k)² = r² ……….(3)
(5 – 2)² + (7 – 3)² = r²
3² + 4² = r²
r² = 25
∴ (3) ⇒ (x – 2)² + (y – 3)² = 25
x² – 4x + 4 + y² – 6y + 9 – 25 = 0
x² + y² – 4x – 6y – 12 = 0

Question 5.
Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter.
Solution:
The equation of a circle with (x₁ , y₁) and (x₂ , y₂ ) as end points of a diameter is
(x – x₁ )(x – x₂) + (y – y₁ )(y – y₂) = 0
Here the end points of a diameter are (3, 4) and (2, -7)
So equation of the circle is (x – 3 )(x – 2 ) + (y – 4) (y + 7.) = 0
x² + y² – 5x + 37 – 22 = 0

Question 6.
Find the equation of the circle through the points (1, 0), (-1, 0) and (0, 1).
Solution:
Let the general equation of the circle be
x² + y² + 2gx + 2fy + c = 0
It passes through the points (1, 0), (-1, 0) and (0,1)
1 + 0 + 2g + c = 0
2g + c = -1 ………(1)
1 + 0 – 2g + c = 0
-2g + c = -1 …………(2)
0 + 1 + 0 + 2f + c = 0
2f + c = -1
(1) + (2) ⇒ 2c = -2
c = -1
substitute in eqn (1)
2g – 1 = -1
2g = 0
g = 0
substitute in eqn (3)
2f – 1 = -1
2f = -1 + 1
2f = 0
f = 0
Therefore, the required equation of the circle
x² + y² – 1 = 0

Question 7.
A circle of area 9π square units has two of its diameters along the lines x + y = 5 and: x – y = 1. Find the equation of the circle.
Solution:

Area of the circle = 9π
(i.e) πr² = 9π
⇒ r² = 9 ⇒ r = 3
(i.e) radius of the circle = r = 3
The two diameters are x + y = 5 and x – y = 1
The point of intersection of the diameter is the centre of the circle = C
To find C: Solving x + y = 5 ……… (1)
x – y = 1 ………. (2)
(1) + (2) ⇒ 2x = 6 ⇒ x = 3
Substituting x = 3 in (1) we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Centre = (3, 2) and radius = 3
So equation of the circle is (x – 3)² + (y – 2)² = 3²
(i.e) x² + y² – 6x – 4y + 4 = 0

Question 8.
If y = 2√2 x + c is a tangent to the circle x² + y² = 16, find the value of c.
Solution:
The condition of the line y = mx + c to be a tangent to the circle x² + y² = a² is
c² = a²( 1 + m²)
a² = 16; m = 2√2 ⇒ m² = 4 × 2 = 8
c² = 16(1+8)
c² = 16(9)
c = ±4 × 3 = ±12
∴ c = ± 12.

Question 9.
Find the equation of the tangent and normal to the circle x² + y² – 6x + 6y – 8 = 0 at (2, 2).
Solution:
The equation of the tangent to the circle x² + y² + 2 gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 9
So the equation of the tangent to the circle
x² + y² – 6x + 6y – 8 = 0 at (x₁, y₁) is
xx₁ + yy₁ – \(\frac{6\left(x+x_{1}\right)}{2}+\frac{6\left(y+y_{1}\right)}{2}\) – 8 = 0
(i.e) xx₁ + yy₁ – 3(x + x₁) + 3(y + y₁) – 8 = 0
Here (x₁, y₁) = (2, 2)
So equation of the tangent is
x(2) + y(2) – 3(x + 2) + 3(y + 2) – 8 = 0
(.i.e) 2x + 2y – 3x – 6 + 3y + 6 – 8 = 0
(i.e) -x + 5y – 8 = 0 or x – 5y + 8=0
Normal is a line ⊥r to the tangent
So equation of normal circle be of the form 5x + y + k = 0
The normal is drawn at (2, 2)
⇒ 10 + 2 + k = 0 ⇒ k = -12
So equation of normal is 5x + y – 12 = 0

Question 10.
Determine whether the points (- 2, 1), (0, 0) and s (- 4, – 3) lie outside, on or inside the circle x² + y² – 5x + 2y – 5 = 0.
Solution:
x² + y² – 5x + 2y – 5 = 0
(i) At (-2, 1) ⇒ (-2)² + 1² – 5(-2) + 2(1) – 5
= 4 + 1 + 10 + 2 – 5 = 12 > 0
∴ (-2, 1) lies outside the circle.

(ii) At(0, 0) ⇒ 0 + 0 – 0 + 0 – 5 = -5 < 0
(0, 0) lies inside the circle.

(iii) At (-4, -3) ⇒ (-4)² + (- 3)² – 5(-4) + 2(-3) – 5 = 16 + 9 + 20 – 6 – 5 = 34 > 0
(-4, -3) lies outside the circle.

Question 11.
Find the centre and radius of the following circles.
(i) x² + (y + 2)² = 0
(ii) x² + y² + 6x – 4y + 4 = 0
(iii) x² + y² – x + 2y – 3 = 0
(iv) 2x² + 2y² – 6x + 4y + 2 = 0
Solution:
(i) x² + (y + 2)² = 0
(i.e) x² + y² + 4y + 4 = 0
Comparing this equation with the general form x² + y² + 2gx + 2fy + c = 0
we get 2g = 0 ⇒ g = 0
2f = 4 ⇒ f= 2 and c = 4
Now centre = (-g, -f) = (0, -2)
Radius = r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{0+4-4}\)
∴ Centre = (0, -2) and radius = 0

(ii) x² + y² + 6x – 4y + 4 = 0
Comparing with the general form we get
2g = 6, 2f = -4
⇒ g = 3, /= -2 and c = 4
Centre = (-g, -f) = (-3, 2)
Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{9+4-4}\)= 3
∴ Centre = (-3, 2) and radius = 3

(iii) x² + y² – x + 2y – 3 = 0
2g = -1; 2f = 2; c = -3
g = \(\frac{-1}{2}\) f = 1
Centre (-g, -f) = (\(\frac{1}{2}\), -1)
Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{\frac{1}{4}+1+3}\)
\(\sqrt{\frac{1+4+12}{4}}\)
\(\sqrt{\frac{17}{4}}\) = \(\frac{\sqrt{17}}{2}\)

(iv) 2x² + 2y² – 6x + 4y + 2 = 0
(÷ by 2) ⇒ x² + y² – 3x + 2y + 1 =0
Comparing this equation with the general form of the circle we get
2g = -3, 2f= 2
g = \(-\frac{3}{2}\), g= 1 and c = 1
So centre = (-g, -f) = (\(\frac{3}{2}\), -1)
and radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}\)
∴ Centre = (\(\frac{3}{2}\), -1) and radius = \(\frac{3}{2}\)

Question 12.
If the equation 3x² + (3 – p) xy + qy² – 2px = 8pq represents a circle, find p and q. Also determine the centre and radius of the circle.
Solution:
3x² + (3 – p) xy + qy² – 2px = 8pq represent a circle means,
Co-efficient of x² = co-efficient of y²
3 = q ⇒ q = 3
Co-efficient of xy = 0
3 – p = 0 ⇒ p = 3
3x² + 3y² – 6x = 8 (3)(3)
3x² + 3y² – 6x – 72 = 0
(÷3) x² + y² – 2x – 24 = 0
2g = -2; 2f = 0; c = -24
g = -1 f = 0
Centre (-g, -f) = (1, 0)
Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{1+0+24}\)
= \(\sqrt{25}\) = 5

Students can download Maths Chapter 2 Integers Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 2 Integers Ex 2.1

Question 1.
Fill in the blanks:
(i) The potable water available at 100 m below the ground level is denoted as ……… m.
(ii) A swimmer dives to a depth of 7 feet from the ground into the swimming pool. The integer that represents this, is ……… feet.
(iii) -46 is to the ……….. of -35 on the number line.
(iv) There are ……… integers from -5 to +5 (both inclusive)
(v) …….. is an integer which is neither positive nor negative.
Solution:
(i) 100
(ii) -7
(iii) left
(iv) 11
(v) 0

Question 2.
Say True or False
(i) Each of the integers -18, 6, -12, 0 is greater than -20.
(ii) -1 is to the right of 0.
(iii) -10 and 10 are at equal distance from 1.
(iv) All negative integers are greater than zero.
(v) All whole numbers are integers.
Solution:
(i) True
(ii) False
(iii) False
(iv) False
(v) True

Question 3.
Mark the numbers 4, -3, 6, -1 and -5 on the number line.
Solution:

Question 4.
On the number line, which number is
(i) 4 units to the right of -7?
(ii) 5 units to the left of 3?
Solution:
(i) -3
(ii) -2

Question 5.
Find the opposite of the following numbers.
(i) 44
(ii) -19
(iii) 0
(iv) -312
(v) 789
Solution:
(i) Opposite of 44 is – 44
(ii) Opposite of-19 is + 19 or 19
(iii) Opposite of 0 is 0
(iv) Opposite of-312 is + 312 or 312
(v) Opposite of 789 is – 789.

Question 6.
If 15 km east of a place is denoted as +15 km, What is the integer that represents 15 km west of it?
Solution:
Opposite of east is west.
∴ If 15 km east is + 15 km, then 15 km west is – 15 km.

Question 7.
From the following number lines, identify the correct and the wrong representations with reason.

Solution:
(i) Wrong, Integers are not continuously marked
(ii) Correct, Integers are correctly marked.
(iii) Wrong, Integer -2 is marked wrongly.
(iv) Correct, Integers are marked at equal distance.
(v) Wrong, negative integers marked wrongly.

Question 8.
Write all the integers between the given numbers.
(i) 7 and 10
(ii) -5 and 4
(iii) -3 and 3
(iv) -5 and 0
Solution:
(i) 8, 9
(ii) -4, -3, -2, -1, 0, 1, 2, 3
(iii) -2, -1, 0, 1, 2
(iv) -4, -3, -2, -1

Question 9.
Put the appropriate signs as <, > or = in the blank.
(i) -7 ___ 8
(ii) -8 ___ -7
(iii) -999 ___ -1000
(iv) 0 ___ -200
Solution:
(i) <
(ii) <
(iii) >
(iv) =
(v) >

Question 10.
Arrange the following integers in ascending order.
(i) -11, 12, -13, 14, -15, 16, -17, 18, -19, -20
(ii) -28, 6, -5, -40, 8, 0, 12, -1, 4, 22
(iii) -100, 10, -1000, 100, 0, -1, 1000, 1, -10
Solution:

(i) -11, 12, -13, 14, -15, 16, -17, 18, -19, -20

• First separating the positive integers 12, 14, 16, 18 and the negative integers -11,-13,-15,-17,-19,-20.
• Then arranging the positive integers in ascending order we get 12, 14, 16, 18 and negative integers in ascending order as -20, -19, -17, -15, -13, -11 4
• Now the ascending order : -20, -19, -17, -15, -13, -11, 12, 14, 16, 18.

(ii) -28, 6, -5, -40, 8, 0, 12, -1, 4, 22

• Positive integers are 6, 8, 12, 4, 22 Negative integers are -28, -5, -40, -1
• Arranging the positive integers in ascending order we get 4, 6, 8, 12, 22 and the negative integers in ascending order -40, -28, -5, -1
• The number 0 is neither positive nor negative and stands in the middle.
• In ascending order : -40, -28, -5, -1, 0, 4, 6, 8,12, 22

(iii) -100, 10, -1000, 100, 0, -1, 1000, 1, -10

• Separating positive integers 10, 100, 1000, 1 and negative integers -100, -1000, -1, -10.
• Now the positive integers in ascending order 1,10,100,1000 and the negative integers in ascending order. -1000, -100, -10, -1
• Also ‘0’ stand in the middle as its is neither positive nor negative.
• ∴ The numbers in ascending order: -1000, -100, -10, -1, 0, 1, 10, 100, 1000.

Question 11.
Arrange the following integers in descending order.
(i) 14, 27, 15, -14, -9, 0, 11, -17
(ii) -99, -120, 65, -46, 78, 400, -600
(iii) 111, -222, 333, -444, 555, -666, 777, -888
Solution:
(i) 27, 15, 14, 11, 0, -9, -14, -17
(ii) 400, 78, 65, -46, -99, -120, -600
(iii) 777, 555, 333, 111, -222, -444, -666, -888

Objective Type Questions

Question 12.
There are ……… positive integers from -5 to 6.
(a) 5
(b) 6
(c) 7
(d) 11
Solution:
(c) 7

Question 13.
The opposite of 20 units to the left of 0 is
(a) 20
(b) 0
(c) -20
(d) 40
Solution:
(a) 20

Question 14.
One unit to the right of -7 is
(a) +1
(b) -8
(c) -7
(d) -6
Solution:
(d) -6

Question 15.
3 units to the left of 1 is
(a) -4
(b) -3
(c) -2
(d) 3
Solution:
(c) -2

Question 16.
The number which determines marking the position of any number to its opposite on a number line is
(a) -1
(b) 0
(c) 1
(d) 10
Solution:
(b) 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4

Question 1.
Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse 2x² + 7y² = 14.
Solution:
2x² + 7y² = 14
(÷ by 14) ⇒ \(\frac{x^{2}}{7}+\frac{y^{2}}{2}\) = 1
comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
we get a² = 7 and b² = 2
The equation of tangent to the above ellipse will be of the form
y = mx + \(\sqrt{a^{2} m^{2}+b^{2}}\) ⇒ y = mx + \(\sqrt{7 m^{2}+2}\)
Here the tangents are drawn from the point (5, 2)
⇒ 2 = 5m + \(\sqrt{7 m^{2}+2}\) ⇒ 2 – 5m = \(\sqrt{7 m^{2}+2}\)
Squaring on both sides we get
(2 – 5m)² = 7m² + 2
25m² + 4 – 20m – 7m² – 2 = 0
18m² – 20m + 2 = 0
(÷ by 2) ⇒ 9m² – 10m + 1 = 0
(9m – 1) (m – 1) = 0
‘ m = 1 (or) m = 1/9
When m = 1, the equation of tangent is
y = x + 3 or x – y + 3 = 0
When m = \(\frac{1}{9}\) the equation of tangent is 9
y = \(=\frac{x}{9}+\sqrt{\frac{7}{81}+2}\) (i.e.) y = \(\frac{x}{9}+\frac{13}{9}\)
9y = x + 13 or x – 9y + 13 = 0

Question 2.
Find the equations of tangents to the hyperbola \(\frac {x^2}{16}\) – \(\frac {y^2}{64}\) = 1 which are parallel to 10x – 3y + 9 = 0.
Solution:
Equation of Hyperbola \(\frac {x^2}{16}\) – \(\frac {y^2}{64}\) = 1
∴ a² = 16, b² = 64
Tangent is parallel to the line
10x – 3y + 9 = 0 is
10x – 3y + k = 0
∴ 3y = 10x + k
y = \(\frac {10}{3}\)x + \(\frac {k}{3}\)
∴ m = \(\frac {10}{3}\) c = \(\frac {k}{3}\)
Condition that the line y = mx + c to be tangent to the hyperbola is
c² = a²m² – b²

k² = 1024
k = ±32
∴ Equation of tangent
⇒ 10x – 3y ± 32 = 0

Question 3.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12. Also find the co-ordinates of the point of contact.
Solution:
The given ellipse is x² + 3y² = 12
(÷ by 12) ⇒ \(\frac{x^{2}}{12}+\frac{y^{2}}{4}\) = 1
(ie.,) Here a² = 12 and b² = 4
The given line is x – y + 4 = 0
(ie.,) y = x + 4
Comparing this line with y = mx + c
We get m = 1 and c = 4
The condition for the line y = mx + c
To be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is c² = a²m² + b²
LHS = c² = 4² = 16
RHS: a²m² + b² = 12( 1 )² + 4 = 16
LHS = RHS The given line is a tangent to the ellipse. Also the point of contact is
\(\left(\frac{-a^{2} m}{c}, \frac{b^{2}}{c}\right)=\left[-\left(\frac{12(1)}{4}\right), \frac{4}{4}\right]\) (i.e.,) (-3, 1)

Question 4.
Find the equation of the tangent to the parabola y² = 16x perpendicular to 2x + 2y + 3 = 0.
Solution:
Equation of the parabola
y² = 16x
4 a = 16
a = 4
Tangent is perpendicular to the line
2x + 2y + 3 = 0 is 2x – 2y + k = 0
2x – 2y + k = 0
2y = 2x + k
y = x + \(\frac {k}{2}\)
m = 1 c = \(\frac {k}{2}\)
Condition that the line y = mx + c to be tangent to the parabola is
c = \(\frac {a}{m}\)
\(\frac {k}{2}\) = \(\frac {4}{1}\)
k = 8
Equation of the tangent
2x – 2y + 8 = 0
÷ by 2 ⇒ x – y + 4 = 0

Question 5.
Find the equation of the tangent at t = 2 to the parabola y² = 8x (Hint: use parametric form).
Solution:
y² = 8x
Comparing this equation with y² = 4ax
we get 4a = 8 ⇒ a = 2
Now, the parametric form for y² = 4ax is x = at², y = 2at
Here a = 2 and t = 2
⇒ x = 2(2)² = 8 and y = 2(2) (2) = 8
So the point is (8, 8)
Now eqution of tangent to y² = 4 ax at (x₁, y₁) is yy₁ = 2a(x + x₁)
Here (x₁, y₁) = (8, 8) and a = 2
So equation of tangent is y(8) = 2(2) (x + 8)
(ie.,) 8y = 4 (x + 8)
(÷ by 4) ⇒ 2y = x + 8 ⇒ x – 2y + 8 = 0
Aliter
The equation of tangent to the parabola y² = 4ax at ‘t’ is
yt = x + at²
Here t = 2 and a = 2
So equation of tangent is
(i.e.,) y(2) = x + 2(2)²
2y = x + 8 ⇒ x – 2y + 8 = 0

Question 6.
Find the equations of the tangent and normal to hyperbola 12x² – 9y² = 108 at θ = \(\frac {π}{3}\) .
(Hint: use parametric form)
Solution:
(i) Equation of the tangent to hyperbola be

⇒ 4x – 3y = 6
⇒ 4x – 3y – 6 = 0

(ii) Equation of the normal to hyperbola be

⇒ 3x + 4y – 42 = 0

Question 7.
Prove that the point of intersection of the tangents at ‘t₁‘ and ‘t₂‘ on the parabola y² = 4ax is [at₁t₂, a(t₁ + t₂)].
Solution:
Equation of the tangent of parabola y² = 4ax be
at t₁ yt₁ = x + at₁² ……….. (1)
at t₂ yt₂ yt = x + at₂² ……….. (2)
(1) – (2) ⇒ y(t₁ – t₂) = a(t₁² – t₂²)
y(t₁ – t₂) = a(t₁ + t₂)(t₁ – t₂)
y = a(t₁ + t₂)
(1) ⇒ t₁a(t₁ + t₂) = x + at₁²
x = at₁² + at₁t₂ – at₁²
x = at₁t₂
Point of intersection be [at₁t₂, a(t₁ + t₂)]

Question 8.
If the normal at the point ‘t₁‘ on the parabola y² = 4ax meets the parabola again at the point t₂ then prove that t₂ = -(t₁ + \(\frac {2}{t_1}\))
Solution:
Equation of normal to y² = 4at’ t’ is y + xt = 2at + at³.
So equation of normal at ‘t₁’ is y + xt₁ = 2at₁ + at₁³
The normal meets the parabola y² = 4ax at ‘t₂’ (ie.,) at (at₂², 2at₂)
⇒ 2at₂ + at₁t₂² = 2at₁ + at₁³
So 2a(t₂ – t₁) = at₁³ – at₁t₂²
⇒ 2a(t₂ – t₁) = at₁(t₁² – t₂²)
⇒ 2(t₂ – t₁) = t₁(t₁ + t₂)(t₁ – t₂)
⇒ 2= -t₁(t₁ + t₂)
⇒ t₁ + t₂ = \(\frac{-2}{t_{1}}\)
⇒ t₂ = \(-t_{1}-\frac{2}{t_{1}}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)