Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6 Textbook Questions and Answers, Notes.

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

Choose 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

x² + y² + x (-5 + 4λ) + y (- 6 + 3λ) + 9 – 19λ = 0

It touches the y-axis put x = 0.

y² + (3λ – 6) y + 9 – 19λ = 0

Now, b² – 4ac = 0

⇒ (3λ – 6)² – 4 (1) (9 – 19λ) = 0

Solving this equation we get

λ = 0 or λ = –\(\frac {40}{9}\)

Question 2.

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distence between the foci is

(a) \(\frac {4}{3}\)

(b) \(\frac {4}{√3}\)

(c) \(\frac {2}{√3}\)

(d) –\(\frac {3}{2}\)

Solution:

(c) \(\frac {2}{√3}\)

Hint:

Length of Latus Rectum \(\frac {2b^2}{a}\) = 8

⇒ b² = 4a …….. (1)

Length of conjugate axes

2b = \(\frac {1}{2}\)(2ae)

⇒ b = \(\frac {1}{2}\) …….. (2)

b² = \(\frac {a^2e^2}{4}\)

Now b² = a²(e² – 1)

\(\frac {a^2e^2}{4}\) = a²(e² – 1)

e² = 4e² – 4

3e² = 4

e² = \(\frac {4}{3}\)

∴ e = \(\frac {2}{√3}\)

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:

C : x² + y² – 4x – 8y – 5 = 0

(x – 2)² + (y – 4)² = 25

C (2, 4); r = 5

Distance from centre < r

-25 < 10 + m < 25

⇒ -25 – 10 < m < 25 – 10

-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).

(a) \(\frac {6}{5}\)

(b) \(\frac {5}{3}\)

(c) \(\frac {10}{3}\)

(d) \(\frac {3}{5}\)

Solution:

(c) \(\frac {10}{3}\)

Hint:

CA = CB

CA² = CB²

(1 – 1)² + (h – 0)² = (1 – 2)² + (h – 3)²

h² = 1 + h² + 9 – 6h

6h = 10

h = \(\frac {10}{6}\) = \(\frac {5}{3}\)

Diameter is 2h = 2(\(\frac {5}{3}\)) = \(\frac {10}{3}\)

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:

Equation of circle

3x² + by² + 4 bx – 6by + b² = 0

a = b ⇒ b = 3

3x² + 3y² + 12x – 18y + 9 = 0

÷ by 3 x² + y² + 4x – 6y + 3 = 0

2g = 4; 2f = -6; c = 3

g = 2; f = -3

r = \(\sqrt {g^2+f^2-c}\)

= \(\sqrt {4+9-3}\)

= \(\sqrt {10}\)

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:

Equation of lines

x² – 8x – 12 = 0

(x – 6)(x – 2) = 0

x = 2, 6

Another lines

y² – 14y + 45 = 0

(y – 5 )(y – 9) = 0

y = 5, 9

Hence the extremities of the diameter are (6, 9) and (2, 5).

Centre is mid point of (6, 9) and (2, 5)

Centre = (\(\frac {6+2}{2}\),\(\frac {9+5}{2}\))

= (4, 7)

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 = 0

(c) 2x + 4y + 3 = 0

(d) x – 2y + 3 = 0

Solution:

(a) x + 2y = 3

Hint:

x² + y² – 2x – 2y + 1 = 0

2g = -2 2f = -2

g = -1 f = -1

Parallel line be 2x + 4y + λ = 0

Centre be (-g, -f) = (1, 1)

Which lies on line

2 + 4 + λ = 0 ⇒ λ = -6

∴ 2x + 4y – 6 = 0 ⇒ x + 2y = 3

Question 8.

If P(x, y) be any point on 16x² + 25y² = 400 with foci F(3, 0) then PF_{1} + PF_{2} is

(a) 8

(b) 6

(c) 10

(d) 12

Solution:

(c) 10

Hint:

16x² + 25y² = 400

\(\frac {x^2}{25}\) + \(\frac {y^2}{16}\) = 1

a² = 25

⇒ ∴ a = ±5

PF_{1} + PF_{2} = major axis = 2a

= 2 × 5 = 10.

Question 9.

The radius of the circle passing through the points (6,2) two of whose diameter are x + y = 6 and x + 2y = 4 is

(a) 10

(b) 2√5

(c) 6

(d) 4

Solution:

(b) 2√5

Hint:

x + y = 6 …….. (1)

x + 2y = 4 ……… (2)

(1) – (2) -y = 2 ⇒ y = -2

(1) ⇒ x – 2 = 6 ⇒ x = 8.

point be (8, -2)

another point (6, 2)

radius = \(\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}\)

= \(\sqrt {(8-6)^2+(2+2)^2}\)

= \(\sqrt {2^2+4^2}\) = \(\sqrt {4+16}\)

= \(\sqrt {20}\) = 2√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:

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:

y² = 4x

4a = 4

a = 1

End points of latus rectum = (a, ±2a)

= (1, ±2)

Normal equation

xyx + 2ay = x_{1}y_{1} + 2 ay_{1}

Equation of normal at points (1, ±2)

y = -x + 3, y = x + 3

x + y – 3 = 0, x – y + 3 = 0

Question 12.

If x + y = k is a normal to the parabola y² = 12x, then the value of k is 14.

(a) 3

(b) -1

(c) 1

(d) 9

Solution:

(d) 9

Hint:

y² = 12x ⇒ 4a = 12

⇒ a = 3

y = mx + c ∴ x + y = k

⇒ y = -x + k

∴ m = -1, c = k.

c = -2am – am² ⇒ k = -2a(-1) – a(-1)³

k = -6(-1) – 3(-1) = 6 + 3 = 9

k = 9

Question 13.

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

(a) \(\frac {√2}{2}\)

(b) \(\frac {√3}{2}\)

(c) \(\frac {1}{2}\)

(d) \(\frac {3}{4}\)

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

(a) (\(\frac {9}{2√2}\), \(\frac {-1}{√2}\))

(b) (\(\frac {-9}{2√2}\), \(\frac {1}{√2}\))

(c) (\(\frac {9}{2√2}\), \(\frac {1}{√2}\))

(d) (3√3, -2√2)

Solution:

(c) (\(\frac {9}{2√2}\), \(\frac {1}{√2}\))

Hint:

a² = 9 b² = 4, 2x – y = 1

y = 2x – 1

m = 2

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

Hint:

a² = 16, b² = 16

(h, k) = (0, 3)

e = \(\sqrt{1-\frac {b^2}{a^2}}\) = \(\sqrt{1-\frac {9}{16}}\)

= \(\sqrt{\frac {7}{16}}\) = \(\frac {√7}{4}\)

ae = 4\(\frac {√7}{4}\) = √7.

F(√7, 0) lies on circle.

(x – h)² + (y – k)² = r²

(√7 – 0)² + (0 – 3)² = r² ⇒ √7² + 3² = r²

7 + 9 = r²

⇒ r² = 16.

∴ (x – 0)² + (y – 3)² = 16

x² + y² – 6y + 9 = 16

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 to

(a) \(\frac {√3}{√2}\)

(b) \(\frac {√3}{2}\)

(c) \(\frac {1}{2}\)

(d) \(\frac {1}{4}\)

Solution:

(d) \(\frac {1}{4}\)

Hint:

ΔOO’A

(1 + y)² = (1 – y)² + 1

1 + y² + 2y = 1 + y² – 2y + 1

4y = 1 ⇒ y = \(\frac {1}{4}\)

Question 17.

Consider an ellipse whose centre is of the origin and its major axis is a long 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:

e = \(\frac {3}{5}\)

2ae = 6 ⇒ 2a(\(\frac {3}{5}\)) = 6

a = 5; b = 4

Area = 4 × \(\frac {1}{2}\) × ab = 2 ab

= 2 × 5 × 4

= 40

Question 18.

Area of the greatest rectangle inscribed in the ellipse \(\frac {x^2}{16}\) + \(\frac {y^2}{9}\) = 1 is

(a) 2ab

(b) ab

(c) \(\sqrt {ab}\)

(d) \(\frac {a}{b}\)

Solution:

(a) 2ab

Hint:

x = a cosθ; y = b sinθ

length = 2acosθ; breadth = 2bsinθ

A = l × b = 4absinθcosθ

A = 2ab sin2θ

\(\frac {dA}{dt}\) = 2ab cos2θ (2)

= 4ab cos2θ

\(\frac {dA}{dt}\) = 0

cos2θ = 0

2θ = \(\frac {π}{2}\)

θ = \(\frac {π}{4}\)

∴ A = 2ab sin²(\(\frac {π}{4}\))

= 2ab sin \(\frac {π}{2}\)

A = 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

(a) \(\frac {1}{√2}\)

(b) \(\frac {1}{2}\)

(c) \(\frac {1}{4}\)

(d) \(\frac {1}{√3}\)

Solution:

(a) \(\frac {1}{√2}\)

Hint:

Distance between foci and end of minor axes = a

∴ F_{1}B = F_{2} = a

F_{1}F_{2} = 2ae

In right angle F_{1}BF_{2}

F_{1}B² + F_{2}B² = F_{1}F_{2}²

a² + a² = (2ae)²

2a² = 4a²e²

e² = \(\frac {2}{4}\) = \(\frac {1}{2}\)

∴ e = \(\frac {1}{√2}\)

Question 20.

The eccentricity of the ellipse

(x – 3)² + (y – 4)² = \(\frac {y²}{9}\) is

(a) \(\frac {√3}{2}\)

(b) \(\frac {1}{3}\)

(c) \(\frac {1}{3√2}\)

(d) \(\frac {1}{√3}\)

Solution:

(b) \(\frac {1}{3}\)

Hint:

PF = e²p³

(x – h)² + (y – k)² = e²(\(\frac {ax+by+c}{\sqrt{a^2+b^2}}\))

(h, k) = (3, 4),

a = 0, c = 0

e² = \(\frac {1}{3}\)

e = \(\frac {1}{3}\)

Question 21.

If the two tangents drawn from a point P to the parabola y2 = 4r are at right angles then the locus of P is {SEA

(a) 2x + 1 = 0

(b) x = -1

(c) 2x – 1 = 0

(d) x = 1

Solution:

(b) x = -1

Hint:

Locus of P = Directrix of y² = 4x; 4a = 4

∴ a = 1

Equation of directrix x = -a = -1

∴ -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:

(x – 3)² + (y – 0)² + λy = 0

At (1, -2), (1 – 3)² + (-2 – 0)² + λy = 0

4 + 4 – 2λ = 0

8 = 2λ

λ = 4

x² – 6x + 9 + y² + 4y = 0

Apply all the point which satisfied that passes through the circle At (+5, -2),

25 – 30 + 9 + 4 – 8 = 0

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:

P(h, k) Q(-2, 0)

x = –\(\frac {9}{2}\) PQ = \(\frac {2}{3}\)

2x + 9 = 0

\(\sqrt {(h+2)^2+k^2}\) = \(\frac {2}{3}\)|\(\frac {2h+9}{2}\)|

(h + 2)² + k² = \(\frac {1}{9}\)(2h + 9)²

h² + 4 + 4h + k² = \(\frac {1}{9}\)(4h² + 36h + 81)

9h² + 36 + 36h + 9k² = 4h² + 36h + 81

5h² + 9k² = 45

\(\frac {h^2}{9}\) + \(\frac {k^2}{5}\) = 1

\(\frac {x^2}{9}\) + \(\frac {y^2}{5}\) = 1

Which is ellipse.

Question 24.

The values of m for which the line y = mx + 2√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:

a² = 9; b² = 16

a = 3; b = 4

c² = a²m² – b²

(2√5)² = 9m² – 16

20 + 16 = 9m²; m² = \(\frac {36}{9}\)

∴ m = 2 which is roots of x² -(a + b)x – 4 = 0

2² -(a + b)2 – 4 = 0

a + b = 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 cordinates of the other end are

(a) (-3, 2)

(b) (2, -5)

(c) (5, -2)

(d) (-2, 5)

Solution:

(a) (-3, 2)

Hint:

2g = -g; 2f = -4

g = -4; f = -2

c(-g, -f) = (4, 2)

\(\frac {x_1+x_2}{9}\) = 4; \(\frac {y_1+y_2}{2}\) = 2

\(\frac {x_1+11}{2}\) = 4; \(\frac {y_1+2}{2}\) = 2

x_{1} = 8 – 11; y_{1} = 4 – 2

x_{1} = -3; y_{1} = 2

∴ Other end be (-3, 2)