Prasanna

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

Question 1.

Solution:

Question 2.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors .
Solution:
Volume of the parallelepiped = \(\| \vec{a}, \vec{b}, \vec{c}]\)

= -264 + 224 + 760 = 720 cubic units

Question 3.
The volume of the parallelepiped whose coterminus edges are , \(-3 \vec{i}+7 \vec{j}+5 \vec{k}\) is 90 cubic units. Find the value of λ
Solution:
Given, Volume of the parallelepiped = 90 cubic units

Question 4.
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4 cubic units, find the value of
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) be the concurrent edges of parallelepiped
Given volume of parallelepiped = 4 cubic units

Question 5.
Find the altitude of a parallelepiped determined by the vectors \(\vec{a}=-2 \hat{i}+5 \hat{j}+3 \hat{k}\), \(\hat{b}=\hat{i}+3 \hat{j}-2 \hat{k}\) and \(\vec{c}=-3 \vec{i}+\vec{j}+4 \vec{k}\) if the base is taken as the parallelogram determined by b and c.
Solution:
Volume = Base Area × Height
\(|[\vec{a}, \vec{b}, \vec{c}]|=|\vec{b} \times \vec{c}|\) × Height

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:

Question 7.
Let If c₁ = 1 and c₂ = 2, find c₃ such that \(\vec{a}, \vec{b}\) and \(\vec{c}\) and c are coplanar.
Solution:

Question 8.
If , show that \([\vec{a} \vec{b} \vec{c}]\) depends neither x nor y.
Solution:

Question 9.
If the vectors

are coplanar, prove that c is the geometric mean of a and b.
Solution:

∴ c is the geometric means of ‘a’ and ‘b’.

Question 10.
Let \(\vec{a}, \vec{b}, \vec{c}\) be three non-zero vectors such that \(\vec{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\) show that .
Solution:

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

Question 1.
If the edges meet a vertex, find the volume of the parallelepiped.
Solution:
Volume of the parallelepiped =
The volume cannot be negative
∴ Volume of parallelepiped = 264 cu. units

Question 2.
If and \(\vec{x} \neq \overrightarrow{0}\) then show that \(\vec{a}, \vec{b}, \vec{c}\) are coplanar.
Solution:

Question 3.
The volume of a parallelepiped whose edges are represented by \(-12 \vec{i}+\lambda k\), \(3 \vec{j}-\vec{k}, 2 \vec{i}+\vec{j}-15 \vec{k}\) is 546. Find the value of λ.
Solution:
Volume of the parallelepiped =
= -12 [-45 + 1] – 0 () + λ [0 – 6] = -12 (-44) -6 λ
= 528 – 6λ = 546 (given)
⇒ -6λ = 546 – 528 = 18
∴ λ = \(\frac{18}{-6}\) = -3

Question 4.
Prove that \(|\vec{a} \vec{b} \vec{c}|\) = abc if and only if \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular.
Solution:

Question 5.
Show that the points (1, 3, 1), (1, 1, -1), (-1, 1, 1), (2, 2, -1) are lying on the same plane.
Solution:

Question 6.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1.
Evaluate the following

Solution:
We know that

(ii)
Solution:

= 0 + 1 + 24 + 4 = 29

Question 2.

Solution:

By using Gamma integral (n = 1; a = α)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6

Question 1.

Solution:

(ii)

Solution:

(iii)

Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:
We know that,

(viii)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.6 Additional Problems

Question 1.
Evaluate

Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

Question 2.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.7

Question 1.
In each of the following cases, determine whether the following function is homogeneous or not. If it is so, find the degree.

Solution:

It is not a homogeneous function

∴ It is a homogeneous function with degree 3.


∴ It is homogeneous function of degree 0.

∴ It is not a homogeneous function

Question 2.
Prove that f(x, y) = x³ – 2x² y + 3xy² + y³ is homogeneous; what is the degree? Verify Euler’s Theorem for f.
Solution:
f (x, y) = x³ – 2x²y + 3xy² + y³
f(tx, ty) = t³x³ – 2(t²x²)(ty) + 3(tx)(t²y²) + t³y³
= t³ [x³ – 2x²y + 3xy² + y³]
f(tx, ty) = t³ f(x, y)
‘f’ is a homogeneous function of degree 3. By Euler’s theorem, we have

∴ Euler’s Theorem verified

Question 3.
Prove that g(x, y) = x log (\(\frac{y}{x}\)) is homogeneous; what is the degree? Verify Euler’s Theorem for g.
Solution:


∴ Euler’s Theorem verified

Question 4.

Solution:

Question 5.

Solution:

∴ ‘f’ is a homogeneous function of degree 1. By Euler’s theorem, we have

Question 6.

Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.7 Additional Problems

Question 1.

Solution:

Question 2.

Solution:
R.H.S. is not a homogeneous and hence

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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:
\(\frac{x^{2}}{16}-\frac{y^{2}}{64}\) = 1
Here a² = 16 and b² = 64
The equation of tangents will be of the form y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)
(i.e.,) y = mx ± \(\sqrt{16 m^{2}-64}\)
Where ‘m’ is the slope of the tangent.
Here we are given that the tangents are parallel to 10x – 3y + 9 = 0
So slope of tangents will be equatl to slope of the given line

Question 3.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12. Also find the coordinates 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:
y² =16x
Comparing this equation with y² = 4ax
we get 4a = 16 ⇒ a = 4
The equation of tangent to the parabola y² = 16x will be of the form

So m = 1
⇒ The equation of tangent will be y = 1(x) + \(\frac{4}{1}\) (i.e.,) y = x + 4
(or) 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 = \(\frac{\pi}{3}\) .
(Hint: Use parametric form)
Solution:
12x² – 9y² = 108

Normal is a line perpendicular to tangent
So equation of normal will be of the form 3x + 4y + k = 0
The normal is drawn at (6, 6)
⇒ 18 + 24 + k = 0 ⇒ k = – 42
So equation of normal is 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 [at₁ t₂, a (t₁ + t₂)].
Solution:
The equation of tangent to y² = 4ax at ‘t’ is given by yt = x + at²
So equation of tangent at ‘t₁‘ is yt₁ = x + at₁²
and equation of tangent at ‘t₂‘ is yt₂ = x + at₂²
To find the point of intersection we have to solve the two equations

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 = \(-\left(t_{1}+\frac{2}{t_{1}}\right)\)
Solution:
Equation of normal to y² =4 at’ 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 = \(-t_{1}-\frac{2}{t_{1}}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)

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

Question 1.
Find the equations of the tangent and normal to the parabolas : x² + 2x – 4y + 4 = 0 at (0, 1)
Solution:

Question 2.
Find the equations of the tangent and normal to the parabola y² = 8x at t = \(\frac{1}{2}\)
Solution:


∴ Equation of the tangent is 2x – y + 1 = 0 and equation of the normal is 2x + 4y – 9 = 0

Question 3.
Find the equations of the tangents: to the parabola y² = 6x, parallel to 3x – 2y + 5 = 0
Solution:


∴ Equation of the tangent is 3x – 2y + 2 = 0

Question 4.
Find the equations of the tangents: to the parabola 4x² – y² = 64 Which are parallel to 10x – 3y + 9 = 0.
Solution:

Question 5.
Find the equation of the tangent from point (2, -3) to the parabola y² = 4x.
Solution:
y² = 4x
Equation of the tangent to the parabola will be of the form y = \(m x+\frac{1}{m}\)
The tangents pass through (2, -3)


The two tangents drawn from (2, -3) are x + y + 1 = 0, x + 2y + 4 = 0

Question 6.
Find the equation of the tangents from the point (2, -3) to the parabola 2x² – 3y² = 6
Solution:
2x² – 3y² = 6

Question 7.
Prove that the line 5x + 12y = 9 touches the hyperbola x² – 9y² = 9 and find the point of contact.
Solution:

(1) = (2) ⇒ The given line is a tangent to the curve i.e., the given line touches the curve. To find the point of contact we have to solve the line and the curve.
The given line 5x + 12y = 9

Question 8.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12. Find the co-ordinates of the point of contact.
Solution:
The given ellipse is x² + 3y² = 12
Given line is x – y + 4 = 0 ⇒ y = x + 4
Here, m = 1 and c = 4
The condition for the line y = mx + c to be a tangent to the ellipse

RHS: a² m² + b = 12(1) + 4 = 16
LHS: RHS ⇒ the given lines is a tangent to the ellipse.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6

Question 1.
If u(x, y) = x²y + 3xy⁴, x = e^t and y = sin t, find \(\frac{d u}{d t}\) and evaluate it at t = 0.
Solution:

= (2xy + 3y⁴) (e^t) + (x² + 12xy³) (cos t)
= (2e^t sin t + 3 sin⁴ t) e^t + [e^2t + 12e^t sin³ t] cos t
= e^t [2e^t sin t + 3 sin⁴ t + e^t (cos t) + 12 sin³t cos t]
at t = 0

Question 2.

Solution:

Question 3.
If w(x, y, z) = x² + y² + z², x = e^t, y = e^t sin t, and z = e^tcos t, find \(\frac{d w}{d t}\)
Solution:
w(x, y, z) = x² + y² + z² ; x = e^t ; y = e^t sin t, z = e^t cos t

Question 4.

Solution:

(Replace x, y, z value)

Question 5.
If w(x, y) = 6x³ – 3xy + 2y², x = e^s, y = cos s, s ∈ R, find \(\frac{d w}{d s}\), and evaluate at s = 0.
Solution:

Question 6.
If z(x, y) = x tan^-1 (x y), x = t², y = s e^t, s, t ∈ R, Find \(\frac{\partial z}{\partial \mathbf{t}}\) and \(\frac{\partial z}{\partial \mathbf{t}}\) at s = t = 1
Solution:
z (x, y) = x tan^-1 (xy) ; x = t² ; y = se^t

Question 7.
Let U (x, y) = e^x sin y, where x = st², y = s²t, s, t ∈ R. Find them at s = t = 1.
Solution:

(Replace x, y values)

Question 8.
Let z(x, y) = x³ – 3x²y³, where x = se^t, y = se^-t, s, t ∈ R. Find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\)
Solution:

Question 9.
W(x,y, z) = xy + yz + zx, x = u -v, y = uv, z = u + v, u, v e R. Find \(\frac{\partial \boldsymbol{w}}{\partial \boldsymbol{u}}\), \(\frac{\partial \boldsymbol{w}}{\partial \boldsymbol{v}}\) them at \(\left(\frac{1}{2}, 1\right)\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6 Additional Problems

Question 1.
Suppose that z = \(y e^{x^{2}}\) where x = 2t and y = 1 – t then find \(\frac{d z}{d t}\).
Solution:


(Since x = 2t and y = 1 – t)

Question 2.
If w = x + 2y + z² and x = cos t ; y = sin t ; z = t. Find \(\frac{d w}{d t}\).
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Question 1.
Evaluate the following:

Solution:

Substitute (2) and (3) in (1), we get

(ii)
Solution:

Dividing both Numerator and Denominator by cos² x

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5 Additional Problems

Question 1.

Solution:

Put 2 tan x, ∴ 2sc² x dx = dt

Question 2.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Evaluate the following:

Question 1.

Solution:
u = x³ ; dv = e^-2
Applying Bernoulli’s formula, we get,

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.4 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.3

Question 1.
Evaluate the following definite integrals :
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

Question 2.
Evaluate the following integrals using properties of integration :
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:

(ix)
Solution:

(x)
Solution:

(xi)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.3 Additional Problems

Question 1.

Solution:
Let f (x) = x³ sin² x = x³ (sin x)²
∴f(- x) = (- x)³ (sin (- x))² = (- x)³ (- sin x)²
= – x³ sin² x = -f(x)
f(-x) = -f(x) ∴ f(x) is an odd function.

Question 2.

Solution:

Question 3.

Solution:
Let f(x) = x sin x
f(-x) = (-x)sin(-x)
= x sin x (∵ sin(-x) = – sin x)
∴ f(x) is an even function

Question 4.

Solution:

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6

Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
The equation of the circle passing through (1,5) and (4, 1) and touching y-axis is x² + y² – 5x – 6y + 9 + λ (4x + 3y – 19) = 0 where λ is equal to …………
(a) 0, \(-\frac{40}{9}\)
(b) 0
(c) \(\frac{40}{9}\)
(d) \(-\frac{40}{9}\)
Solution:
(a) 0, \(-\frac{40}{9}\)
Hint:
x² + y² – 5x – 6y + 9 + λ (4x + 3y -19) = 0
ie., x² + y² + x(4λ – 5) + y(3λ – 6) + 9 — 19λ = 0
Since it touches y axis x = 0
⇒ y² + y(3λ – 6) + 9 – 19λ = 0
It is a quadratic in y ,
Since the circle touches y axis the roots must be equal ⇒ b² – 4ac = 0
ie.,(3λ – 6)² – 4(1) (9 – 19λ) = 0
Solving we get λ = o or \(-\frac{40}{9}\)

Question 2.
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is ………..

Solution:
(c) \(\frac{2}{\sqrt{3}}\)
Hint:
Given \(\frac{2 b^{2}}{a}\) = 8 and 2b = ae

Question 3.
The circle x² + y² = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if …………..
(a) 15 < m < 65
(b) 35 < m < 85
(c) -85 < m < -35
(d) -35 < m < 15
Solution:
(d) -35 < m < 15
Hint:
x² + y² – 4x – 8y – 5 =0
3x – 4y = m Solving (1) and (2) we get
from (2) ⇒ 3x = 4y + m
x = \(\frac{4 y+m}{3}\)
Substituting x value in (1) we get
\(\left(\frac{4 y+m}{3}\right)^{2}\) + y² – \(4\left(\frac{4 y+m}{3}\right)\) – 8y – 5 = 0
It is a quadratic in y and given that the roots are distinct
⇒ b² – 4ac > 0
On simplifying we get
⇒ – 9m² – 18w + 4725 >0 ⇒ m² + 20m – 525 < 0
⇒ (m + 35) (m – 15) ≤ 0 ⇒ m lies between -35 and 15 ie„ – 35 < m < 15

Question 4.
The length of the diameter of the circle which touches the x -axis at the point (1,0) and passes through the point (2, 3) …………

Solution:
(c) \(\frac{10}{3}\)
Hint:
Since radius is ⊥ r to tangent it passes through the centre (1, y)
AC = CB = radius
AC² = CB² ⇒ y² = 1 + (y – 3)²
6y = 10 ⇒ y = \(\frac{5}{3}\)
Diameter 2y = \(\frac{10}{3}\)

Question 5.
The radius of the circle 3x² + by² + 4bx – 6by + b² = 0 is ……………
(a) 1
(b) 3
(c) \(\sqrt{10}\)
(d) \(\sqrt{11}\)
Solution:
(c) \(\sqrt{10}\)
Hint:
Here b = 3
3x² + 3y² + 12x- 18y + 9 = 0
(÷ 3) zz> x² + y² + 4x – 6y + 3 = 0
Comparing with general form
g = 2,f = -3, c = 3
∴ radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+9-3}=\sqrt{10}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is ………..
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)

Solution:
(a) (4, 7)
Hint:
(x – 6) (x – 2) = 0 ⇒ x = 2 or 6
(y – 9) (y – 5) = 0 ⇒ y = 5 or 9

Question 7.
The equation of the normal to the circle x² + y² – 2x – 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is …………..
(a) x + 2y = 3
(b) x + 2y + 3 = 3
(c) 2x + 4y + 3 = 0
(d) x – 2y + 3 = 0
Solution:
(a) x + 2y = 3
Hint:
Centre of the circle = (1, 1) and radius = r = 1
Normal is parallel to 2x + 4y = 3
So equation of normal will be the form 2x + 4y = k
It passes throug the centre (1, 1) ⇒ 2 + 4 = 6 = k
So equation of the normal is x + 4y = 6
(÷ by 2) x + 2y = 3

Question 8.
If P(x, y) be any point on 16x² + 25y² = 400 with foci F₁(3, 0) and F₂(- 3, 0) then PF₁ + PF₂ is
(a) 8
(b) 6
(c) 10
(d) 12
Solution:
(c) 10
Hint:
F₁P + F₂P = 2a
Here the equation is 16x² + 25y² = 400
(÷ by 400) ⇒ \(\frac{16 x^{2}}{400}+\frac{25 y^{2}}{400}\) = 1 ⇒ \(\frac{x^{2}}{25}+\frac{y^{2}}{16}\) = 1
a² = 25 => a = 5
∴ 2a = 10

Question 9.
The radius of the circle passing through the point (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is …………..
(a) 10
(b) \(2 \sqrt{5}\)
(c) 6
(d) 4
Solution:
(b) \(2 \sqrt{5}\)
Hint:
The point of intersection of the diameters is the centre.
Now solving x + y = 6 …………. (1)
x + 2y = 4 ………….. (2)
(1) – (2) ⇒ -y = 2 ⇒ y = -2
Substituting y = -6 in (1)
x – 2 = 6 ⇒ x = 8
Centre = (8, -2)
The circle passes through (6, 2)
∴ radius = \(\sqrt{(8-6)^{2}+(-2-2)^{2}}=\sqrt{4+16}\)
= \(\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5}\)

Question 10.
The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = -1 is ……………
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac{1}{2}\)(a² + b²)
Solution:
(b) 2(a² + b²)
Hint:
The foci are (± ae, 0) and (0, ± ae)

The above diagram is a rhombus.
Its area = \(\frac{1}{2}\)d₁d₂
= \(\frac{1}{2}\) (2ae) (2ae) = 2a² e²
But we know b² = a² (e² – 1)
i.e., b² = a² e² – a²
⇒ b² + a² = a² e²
So area = 2(a² + b²)

Question 11.
If the normals of the parabola y² = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)² + (y + 2)² = r² , then the value of r² is ………………..
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(a) 2
Hint:
The normals are x ± y = 3
Distance from (3, -2) on both normal is r
(i.e.,) r = \(\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}\) ⇒ r² = 2

Question 12.
If x + y = k is a normal to the parabola y² = 12x, then the value of k is ……………
(a) 3
(b) -1
(c) 1
(d) 9
Solution:
(d) 9
Hint:
Slope of normal = -1 and Slope of tangent = -1
The condition for y = mx + c to be a tangent to the parabola
y² = 4ax is c = \(\frac{a}{m}\)
Here the parabola is y² = 12x ⇒ a = 3 and m = 1
∴ c = \(\frac{3}{1}\) = 3
∴ Equation of tangent is y = x + 3
Solving the parabola and tangent we get the points (3, 6) and (3,-6)
∴ k = -3 or k = 9

Question 13.
The ellipse E₁ : \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E₂ passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse is …………….

Solution:
(c) \(\frac{1}{2}\)
Hint:

Question 14.
Tangents are drawn to the hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{4}\) = 1 parallel to the straight line 2x – y = 1. One of the points of contact of tangents on the hyperbola is ……………

Solution:
(c) \(\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
Hint:
The tangents are parallel to 2x – y = 1
So equation of tangents will be of the form
2x – y = k ⇒ y = 2x – k
Comparing this equation with y = mx + c we get m = 2 and c = -k
The given hyperbola is \(\frac{x^{2}}{9}-\frac{y^{2}}{4}\) = 1
⇒ a² = 9 and b² = 4
Now the condition for the line y = mx + c to be a tangent to

Question 15.
The equation of the circle passing through the foci of the ellipse \(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1 having centre at (0, 3) is ……………….
(a) x² + y² – 6y – 7 = 0
(b) x² + y² – 6y + 7 = 0
(c) x² + y² – 6y – 5 = 0
(d) x² + y² – 6y + 5 = 0
Solution:
(a) x² + y² – 6y – 7 = 0

The distance between centre and foci = radius of the circle
ie., r = \(\sqrt{(0 \pm \sqrt{7})^{2}+(3-0)^{2}}=\sqrt{7+9}=\sqrt{16}\) = 4
Now Centre = (0, 3); r = 4
Equation of the circle is (x – 0)2 + (y -3)2 = 42
ie., x² + y² – 6y + 9 – 16 = 0
x² + y² – 6y – 7 = 0

Question 16.
Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centered at (0, y) passing through the origin and touching the circle C externally, then the radius of T is equal ……………..

Solution:
(d) \(\frac{1}{4}\)
Hint:
Equation of circle is x² + y² + 2gx + 2fy + c = 0
i.e., (1 + y²)² = (1 – y)² = 1
Solving we get y = \(\frac{1}{4}\)

Question 17.
Consider an ellipse whose centre is of the origin and its major axis is along x-axis. If its eccentricity is \(\frac{3}{5}\) and the distance between its foci is 6, then the area of the quadrilateral inscribed in the ellipse with diagonals as major and minor axis of the ellipse is ……………
(a) 8
(b) 32
(c) 80
(d) 40
Solution:
(d) 40
Hint:

Question 18.
Area of the greatest rectangle inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is …………….
(a) 2ab
(b) ab
(c) \(\sqrt{a b}\)
(d) \(\frac{a}{b}\)
Solution:
(a) 2ab
Hint:
Area of the greatest rectangle is (\(\sqrt{2}\) a) (\(\sqrt{2}\) b) = 2ab

Question 19.
An ellipse has OB as semi minor axes, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is …………..

Solution:
(a) \(\frac{1}{\sqrt{2}}\)
Hint:

Question 20.
The eccentricity of the ellipse (x – 3)² + (y – 4)² = \(\frac{y^{2}}{9}\) is ……………

Solution:
(b) \(\frac{1}{3}\)
Hint:
The given equation is of the form FP² = e²PM²
i.e., (x – 3)² + (y – 4)² = \(\frac{1}{9}\)(y²)
⇒ e² = \(\frac{1}{9}\) ⇒ e = \(\frac{1}{3}\)

Question 21.
If the two tangents drawn from a point P to the parabola y² = 4x are at right angles then the locus of P is ………………….
(a) 2x + 1 = 0
(b) x = -1
(c) 2x – 1 = 0
(d) x = 1
Solution:
(b) x = -1
Hint:
When the tangents at drawns from a point on the directrix are at right angles.
So equation of directrix to y² = 4x is x = -1

Question 22.
The circle passing through (1, -2) and touching the axis of x at (3, 0) passing through the point …………..
(a) (-5, 2)
(b) (2, -5)
(c) (5, -2)
(d) (-2, 5)
Solution:
(c) (5, -2)
Hint:

Let the centre be (3, h)
r = k
Equation of the circle is
(x – 3)² + (y – h)² = r² = (k)²
It passes through (1, -2) ⇒ k² = 8
Substituting (5, -2) the equation satisfies.

Question 23.
The locus of a point whose distance from (-2, 0) is \(\frac{2}{3}\) times its distance from the line x = \(\frac{-9}{2}\) is ……….
(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle
Solution:
(c) an ellipse
Hint:
Here e = \(\frac{2}{3}\) < 1
So the conic is an ellipse

Question 24.
The values of m for which the line y = mx + 2\(\sqrt{5}\) touches the hyperbola 16x² – 9y² = 144 are the roots of x² – (a + b)x -4 = 0, then the value of (a + b) is
(a) 2
(b) 4
(c) 0
(d) -2
Solution:
(c) 0
Hint:

c² = a² m² – b²
Here the given line is y = mx + 2 \(\sqrt{5}\)
⇒ m = m and c = 2\(\sqrt{5}\)
The condition is c² = a²m² – b²
ie., (2\(\sqrt{5}\))²= 9(m²)- 16
⇒ 9m² = 20 + 16 = 36 ⇒ m² = 4
i.e., m = ± 2
a = 2, b = -2 (say)
So a + b = 2 – 2 = 0

Question 25.
If the coordinates at one end of a diameter of the circle x² + y² – 8x – 4y + c = 0 are (11, 2), the coordinates of the other end are ……………
(a) (-5, 2)
(b) (-3, 2)
(c) (5, -2)
(d) (-2, 5)
Solution:
(b) (-3, 2)
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6 Additional Problems

Choose the most appropriate answer:
Question 1.
The parabola y² = 4ax passes through the point (2, -6), the the length of its latus rectum is …………..
(a) 9
(b) 16
(c) 18
(d) 6
Solution:
(c) 18
Hint:
The given parabola is y² = 4ax
Since it passes through (2, -6)
∴ (-6)² = 4a(2)
⇒ 36 = 8a a = \(\frac{36}{8}=\frac{9}{2}\)
Length of the latus rectum = 4a = \(4\left(\frac{9}{2}\right)\) = 18

Question 2.
The vertex of the parabola x² + 12x – 9y = 0 is …………
(a) (6, -1)
(b) (-6, 4)
(c) (6, 4)
(d) (-6, -4)
Solution:
(d) (-6, -4)
Hint:
Given parabola is x² + 12x – 9y = 0
⇒ (x² + 12x + 36) – 36 – 9y = 0
⇒ (x + 6)² = 9y + 36 ⇒ (x + 6)² = 9 (y + 4)
∴ Vertex = (-6, -4)

Question 3.
The length of the major axis of an ellipse is three times the length of minor axis, its eccentricity is ……………

Solution:
(d) \(\frac{2 \sqrt{2}}{3}\)
Hint:
Length of major axis = 2a and Length of minor axis = 2b
2a = 3(2b) ⇒ a = 3b

Question 4.
The eccentricity of the ellipse 9x² + 4y² = 36 is …………..

Solution:
(d) \(\frac{\sqrt{5}}{3}\)
Hint:
Given equation of the ellipse 9x² + 4y² = 36
On dividing both the sides by 36, we get

Question 5.
S and T are the foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is …………..

Solution:
(c) \(\frac{1}{2}\)
Hint:

Question 6.
The sum of the focal distances from any point of the ellipse 9x² + 16y² = 144 is …………….
(a) 32
(b) 18
(c) 16
(d) 8
Solution:
(d) 8
Hint:
We have, 9x² + 16y² = 144
⇒ \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1 (Dividing both sides by 144)
Here a² = 16, b² = 9 ⇒ a = 4, b = 3
Length of major axis = 2a = 2(4) = 8
Since the sum of the focal distances from any point on the ellipse is equal to its major axis
∴ Required sum = 8

Question 7.
If the eccentricities of two ellipse \(\frac{x^{2}}{169}+\frac{y^{2}}{25}\) = 1 and \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are equal then \(\frac{a}{b}\) = ………………

Solution:
(c) \(\frac{13}{5}\)
Hint:

Question 8.
Equation of the hyperbola, whose eccentricity \(\frac{3}{2}\) and foci at (±2, 0) is ……………

Solution:
(a) \(\frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}\)
Hint:
Eccentricity e = \(\frac{c}{a}=\frac{3}{2}\)
Foci are (± c, 0) = (± 2, 0) (Given c = 2)

Question 9.
If e₁ is the eccentricity of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1 and if e₂ is the eccentricity of the hyperbola 9x² – 16y² = 144, then e₁e₂ is …………….
(a) \(\frac{16}{25}\)
(b) 1
(c) Greater than 1
(d) Less than \(\frac{1}{2}\)
Solution:
(b) 1
Hint:

Question 10.
The point of intersection of the tangents at t₁ = t and t₂ = 3 t to the parabola y² = 8x is ………….
(a) (6t², 8t)
(b) (8t, 6t²)
(c) (t², 4t)
(d) (4t, t²)
Solution:
(a) (6t², 8t)
Hint:
Point of intersection of the tangents at t₁ and t₂ to y² = 4ax is [a t₁ t₂, a(t₁ + t₂)]
Here, t₁ = t, t₂ = 3t, a = latex]\frac{8}{4}[/latex] = 2
So a t₁ t₂ = 2(t) (3t) = 6t²
a(t₁ + t₂) = 2(t + 3t) = 8t
∴ Point = (6t², 8t²)