Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.2 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.2

Question 1.

Find the equation of the parabola in each of the cases given below:

(i) focus (4, 0) and directrix x = -4.

(ii) passes through (2, – 3) and symmetric about y axis.

(iii) vertex (1,-2) and focus (4, – 2).

(iv) end points of latus rectum (4, -8) and (4, 8).

Solution:

(i) focus (4, 0) and directrix x = -4

Parabola is open rightwards vertex (0, 0)

a = 4

(Distance AS = 4 unit)

f² = 4(4) x

Equation of parabola

y² = 16x.

(ii) passes through (2, – 3) and symmetric about y-axis

x^{2} = 4ay

It passes through (2, -3)

⇒ 2^{2} = 4a(-3)

4 = -12a ⇒ a = \(-\frac{1}{3}\) ⇒ 4a = \(-\frac{4}{3}\)

∴ Equation of parabola is x^{2} = \(-\frac{4}{3}\) y

3x^{2}= – 4y.

(iii) vertex (1,-2) and focus (4, – 2)

In given data the parabola is open rightwards and symmetric about the line parallel to x-axis.

Equation of parabola

(y – k)² = 4a(x – h)

Vertex (h, k) = (1, -2)

(y + 2)² = 4a(x – 1)

a = AS = 3

Equation of parabola

(y + 2)² = 4(3)(x – 1)

(y + 2)² = 12(x – 1)

(iv) end points of latus rectum (4, -8) and (4, 8)

Focus = (4, 0)

Equation of the parabola will be of the form y^{2} = 4ax

Here a = 4

⇒ y^{2} = 16x

Question 2.

Find the equation of the ellipse in each of the cases given below:

(i) foci (± 3, 0), e = \(\frac {1}{2}\)

(ii) foci (0, ±4) and end points of major axis are (0, ±5).

(iii) length of latus rectum 8, eccentricity = \(\frac {3}{5}\) and major axis on x-axis.

(iv) length of latus rectum 4, distance between foci and major axis as y axis.

Solution:

(i) foci (± 3, 0), e = \(\frac {1}{2}\)

foci (± c, 0) = (± 3, 0)

e = \(\frac {1}{2}\)

c = ae = 3

a(\(\frac {1}{2}\)) = 3

a = 6 ⇒ a² = 36

b² = a² – c²

b² = 36 – 9 = 27

b² = 27

Equation of the ellipse be \(\frac {x^2}{a^2}\) + \(\frac {y^2}{b^2}\) = 1

\(\frac {x^2}{36}\) + \(\frac {y^2}{27}\) = 1

(ii) foci (0, ±4) and end points of major axis are (0, ±5)

foci (0, ±c) = (0, +4)

vertex (0, ±a) = (0, ±5)

∴ c = 4, a = 5

ae = 4

5e = 4

e = \(\frac {4}{5}\)

b² = a² – c²

= 25 – 16

b² = 9

Equation of the ellipse be \(\frac {x^2}{b^2}\) + \(\frac {y^2}{a^2}\) = 1

\(\frac {x^2}{9}\) + \(\frac {y^2}{25}\) = 1

(iii) length of latus rectum 8, eccentricity = \(\frac {3}{5}\) and major axis on x-axis.

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

Latus rectum \(\frac {2b^2}{a}\) = 8

(iv) length of latus rectum 4, distance between foci 4√2 and major axis as y-axis

Given \(\frac{2 b^{2}}{a}\) = 4 and 2ae = \(4 \sqrt{2}\)

Now \(\frac{2 b^{2}}{a}\) = 4 2b^{2} = 4a

⇒ b^{2} = 2a

2ae = \(4 \sqrt{2}\) ae = \(2 \sqrt{2}\)

So a^{2}e^{2} = 4(2) = 8

We know b^{2} = a^{2}(1 – e^{2}) = a^{2} – a^{2}e^{2}

⇒ 2a = a^{2} – 8 ⇒ a^{2} – 2a -8 = 0

⇒ (a – 4) (a +2) = 0 ⇒a = 4 or -2

As a cannot be negative

a = 4 So a^{2} = 16 and b^{2} = 2(4) = 8

Also major axis is along j-axis

So equation of ellipse is \(\frac{x^{2}}{8}+\frac{y^{2}}{16}\) = 1

Question 3.

Find the equation of the hyperbola in each of the cases given below:

(i) foci (± 2, 0), eccentricity = \(\frac {3}{2}\)

(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.

(iii) passing through (5, -2) and length of the transverse axis along x axis and length 8 units.

Solution:

(i) foci (± 2, 0), eccentricity = \(\frac {3}{2}\)

(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.

Distance CS = ae = 6 …….. (1)

Directrix \(\frac {a}{e}\) = 4 ……… (2)

(1) × (2) ⇒ ae × \(\frac {a}{e}\) = 24

a² = 24

∴ c = ae = 6

b² = c² – a²

= 36 – 24 = 12

The transverse axis is parallel to x-axis

∴ \(\frac {(x-h)^2}{a^2}\) – \(\frac {(y – k)^2}{b^2}\) = 1 (h, k) = (2, 1)

\(\frac {(x-2)^2}{24}\) – \(\frac {(y – 1)^2}{12}\) = 1

(iii) passing through (5,-2) and length of the transverse axis along x axis and of length 8 units.

Transverse axis along x-axis

\(\frac {x^2}{a^2}\) – \(\frac {y^2}{b^2}\) = 1

Length of transverse axis 2a = 8

⇒ a = 4

Question 4.

Find the vertex, focus, equation of directrix and length of the latus rectum of the following:

(i) y² = 16x

(ii) x² = 24y

(iii) y² = -8x

(iv) x² – 2x + 8y + 17 = 0

(v) y² – 4y – 8x + 12 = 0

Solution:

(i) y² = 16x

4a = 16

a = 4

(a) Vertex V (0, 0)

(b) Focus S(a, 0) = S(4, 0)

(c) Equation of the directrix x = – a

x = -4 ⇒ x + 4 = 0

(d) Length of the latus rectum = 4a = 4(4)

= 16

(ii) x² = 24y

(a) Vertex V (0, 0),

(b) Focus S (0, a) = S(0, 6)

(c) Equation of the directrix y = -a = -6

⇒ y + 6 = 0

(d) Length of the latus rectum = 4a

= 4 (6) = 24

(iii) y² = -8x

4a = 8,

a = 2

(a) Vertex V(0, 0) = ( 0, 0)

(b) Focus S(-a, 0) = (-2, 0)

(c) Equation of the directrix x = a = 2

x – 2 = 0

(d) Length of the latus rectum 4a = 8

(iv) x² – 2x + 8y + 17 = 0

x² – 2x = -8y – 17

(x – 1)² = -8y – 17 + 1

(x – 1)² = -8y – 16

(x – 1)² = -8(y + 2)

It is form of (x – h)² = -4a(y – k)

4a = 8 ⇒ a = 2

(a) Vertex be (h, k) = (1, -2)

(b) Foeus = (0 +h, -a + k) = (0 + 1, -2 – 2) = (1, -4)

(c) Equation of the directrix is y + k + a = 0

y – 2 + 2= 0

y = 0

(d) Length of latus rectum is 4a = 4 × 2 = 8 units,

(v) y² – 4y – 8x + 12 = 0

y² – 4y = 8x – 12

(y – 2)² = 8x – 12 + 4

= 8x – 8

= 8 (x – 1)

(y – 2)² = 8 (x – 1)

It is form of (y – k)² = Aa(x – h)

4a = 8 ⇒ a = 2

(a) Vertex (h, k) = (1, 2)

(b) Focus = (a+h, 0 + k) = (2 + 1, 0 + 2) = (3, 2)

(c) Equation of the directrix x = -a + h

= -2 + 1

= -1

x + 1 = 0

(d) Length of latus rectum is

4a = 4 × 2 = 8 units.

Question 5.

Identify the type of conic and find centre, foci, vertices and directrices of each of the following:

(i) \(\frac {x^2}{25}\) + \(\frac {y^2}{9}\) = 1

(ii) \(\frac {x^2}{3}\) + \(\frac {y^2}{10}\) = 1

(iii) \(\frac {x^2}{25}\) – \(\frac {y^2}{144}\) = 1

(iv) \(\frac {y^2}{16}\) – \(\frac {x^2}{9}\) = 1

Solution:

(i) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1

It is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, which is an ellipse

Here a^{2} = 25, b^{2} = 9

a = 5, b = 3

e^{2} = \(\frac{a^{2}-b^{2}}{a^{2}}=\frac{25-9}{25}=\frac{16}{25}\) ⇒ e = \(\frac{4}{5}\)

Now e = \(\frac{4}{5}\) and a = 5 ⇒ ae = 4 and \(\frac{a}{e}=\frac{5}{4 / 5}=\frac{25}{4}\)

Here the major axis is along x axis

∴ Centre = (0, 0)

Foci = (± ae, 0) = (± 4, 0)

Vertices = (± a, 0) = (±5, 0)

Equation of directrix x = ± \(\frac{a}{e}\) (ie.,) x = ± \(\frac{25}{4}\)

(ii) \(\frac {x^2}{3}\) + \(\frac {y^2}{10}\) = 1

It is an ellipse. The major axis is along y axis

a² = 10 b² = 3

a = \(\sqrt {10}\) b = √3

c² = a² – b²

= 10 – 3 = 7

c = √7

ae = √7 .

\(\sqrt {10}\) = √7

e = \(\sqrt {\frac{7}{10}}\)

(a) Centre (0, 0)

(b) Vertex (0, ±a) = (0, ±\(\sqrt {10}\))

(c) Foci (0, ±c) – (0, ±√7)

(d) Equation of the directrix a

y = ±\(\frac{a}{e}\)

= ±\(\frac{\sqrt {10}}{√7}\).\(\sqrt {10}\) = ±\(\frac {10}{√7}\)

y = ±\(\frac {10}{√7}\)

(iii) \(\frac {x^2}{25}\) – \(\frac {y^2}{144}\) = 1

It is Hyperbola. The transverse axis the x axis.

a² = 25; b² = 144

a = 5; b = 12

c² = a² + b²

= 25 + 144 = 169

c = 13

ae = 13

5e = 13

e = \(\frac {13}{5}\)

(a) Centre (0, 0)

(b) Vertex (± a, 0) = (± 5, 0)

(c) Foci (± c, 0) = (± 13, 0)

(d) Equation of the directrix

x = ±\(\frac {a}{e}\) = ±\(\frac {5}{\frac{13}{5}}\) = ±\(\frac {25}{13}\)

x = ±\(\frac {25}{13}\)

(iv) \(\frac {y^2}{16}\) – \(\frac {x^2}{9}\) = 1

It is Hyperbola. The transverse axis the y axis.

a² = 16; b² = 9

a = 4; b = 3

c² = a² + b²

= 16 + 6 = 25

c = 5

ae = 5

4e = 5

e = \(\frac {5}{4}\)

(a) Centre (0, 0)

(b) Vertex (0, ±a) = (0, ±4)

(c) Foci (0, ±ae) = (0, ±5)

(d) Equation of the directrix

y = ±\(\frac {a}{e}\) = ±\(\frac {4}{\frac{5}{4}}\) = ±\(\frac {16}{5}\)

y = ±\(\frac {16}{5}\)

Question 6.

Prove that the length of the latns rectum of the hyperbola \(\frac {x^2}{a^2}\) – \(\frac {y^2}{b^2}\) = 1 is \(\frac {2b^2}{a}\)

Solution:

The latus rectum LL’ of an hyperbola \(\frac {x^2}{a^2}\) – \(\frac {y^2}{b^2}\) = 1 passes through S(ae, 0)

Hence L is (ae, y_{1})

Hence proved

Question 7.

Show that the absolute value of the difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.

Solution:

Let P be a point on the hyperbola.

Definition of conic

\(\frac {SP}{PM}\) = e; \(\frac {S’P}{PM’}\) = e

SP = e(PM) ……..(1)

S’P = e (PM’) ……….(2)

(2) – (1) ⇒ S’P – SP = e PM’- e PM

= e(PM’ – PM)

= e MM’

= e ZZ’

[∵ MM’ = ZZ’ = \(\frac {2a}{e}\) ]

= e(\(\frac {2a}{e}\))

S’P – SP = 2a (constant)

= length of the transverse axis.

Question 8.

Identify the type of conic and find centre, foci, vertices and directrices of each of the following:

(v) 18x² + 12y² – 144x + 48y + 120 = 0

(vi) 9x² – y² – 36x – 6y + 18 = 0

Solution:

(i) \(\frac {(x-3)^2}{225}\) + \(\frac {(y-4)^2}{289}\) = 1

It is an ellipse. The major axis is parallel to y axis

a² = 289, b² = 225

a = 17, b = 15

c² = a² – b²

= 289 – 225 = 64

c = 8

ae = 8

17e = 8

e = \(\frac {8}{17}\)

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

Vertices (h, ±a + k) = (3, 17 + 4) & (3, -17 + 4)!

= (3, 21) and (3, -13)

Foci (h + 0, ± c + k) = (3, 8 + 4) & (3, -8 + 4)

= (3, 12) and (3, -4)

(ii) \(\frac {(x+1)^2}{100}\) + \(\frac {(y-2)^2}{64}\) = 1

It is an ellipse. The major axis is parallel to the x-axis.

a² = 100, b² = 64

a = 10, b = 8

c² = a² – b²

= 100 – 64 = 36

c = 6

ae = 6

10 e = 6

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

centre (h, k) = (-1, 2)

vertices (h±a, k) = (-1±10, 2)

= (-1+10, 2) & (-1-10, 2)

= (9, 2) & (-11, 2)

foci (h±c, k) = (-1±6, 2)

= (-1+6, 2) & (-1-6, 2)

= (5, 2) & (-7, 2)

directrix x = ±\(\frac {a}{e}\) + h

(iii) \(\frac {(x+3)^2}{225}\) + \(\frac {(y-4)^2}{64}\) = 1

It is an hyperbola. The transverse axis is parallell to x axis.

a² = 225, b² = 64

a = 15, b = 8

c² = a² – b²

= 225 + 64

c² = 289

c = 17

ae = 17

5e = 17

e = \(\frac {17}{15}\)

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

vertices (h±a, k) = (-3±15, 4)

= (-3+15, 4) & (-3-15, 4)

= (12, 4) & (-18, 4)

foci (h±c, k) = (-3±17, 4)

= (-3+17, 4) & (-3-17, 4)

= (14, 4) & (-20, 4)

directrix x = ±\(\frac {a}{e}\) + h

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

It is an hyperbola. The transverse axis is parallell to y axis.

a² = 25, b² = 16

a = ±5, b = 4

c² = a² + b²

= 25 + 16

= 41

c = \(\sqrt {41}\)

ae = \(\sqrt {41}\)

5 e = \(\sqrt {41}\)

e = \(\frac {\sqrt {41}}{5}\)

centre (h, k) = (-1, 2)

vertices (h, ±a + k) = (-1, ±5 + 2)

= (-1, 5 + 2) & (-1, -5 + 2)

= (-1, 7) & (-1, -3)

foci (h, ±c + k) = (-1, ±\(\sqrt {41}\) + 2)

= (-1, \(\sqrt {41}\) + 2) & (-1, –\(\sqrt {41}\) + 2)

directrix x = ±\(\frac {a}{e}\) + k

(v) 18x² + 12y² – 144x + 48y + 120 = 0

18x² – 144x + 12y² + 48y = -120

18 (x² – 8x) + 12 (y² + 4y) = -120

18 (x – 4)² + 12 (y + 2)² = -120 + 288 + 48

18(x – 4)² + 12 (y + 2)² = 216

\(\frac {18(x-4)^2}{216}\) + \(\frac {12(y+2)^2}{216}\) = 1

\(\frac {(x-4)^2}{12}\) + \(\frac {(y+2)^2}{18}\) = 1

It is an ellipse. The major axis is parallell to y axis.

a² = 18, b² = 12

a = \(\sqrt {18}\), b = 2√3

= 3 √2

c² = a² + b²

= 18 – 12

= 6

c = √6

ae = √6

3√2 e = √6

e = \(\frac {√6}{3√2}\) = \(\frac {√3}{3}\) = \(\frac {1}{√3}\)

centre (h, k) = (4, -2)

vertices (h, ±a + k) = (4, ±3√2 – 2)

= (4, 3√2 – 2) & (4, -3√2 – 2)

foci (h, ±c + k) = (4, ±√6 – 2)

= (4, ±√6 – 2) & (4, -√6 – 2)

directrix y = ±\(\frac {a}{e}\) + k

= ±\(\frac {3√2}{1}\) – 2

= ±3√6 – 2

y = 3√6 – 2 & y = -3√6 – 2

(vi) 9x² – y² – 36x – 6y + 18 = 0

9x² – 36x – y² – 6y = – 18

9(x² – 4x) – (y² + 6y) = -18

9(x – 2)² – (y + 3)² = -18 + 36 – 9

9(x – 2)² – (y + 3)² = 9

\(\frac {(x-2)^2}{1}\) – \(\frac {(y+3)^2}{9}\) = 1

It is hyperbola. The transverse axis is parallel to x axis.

a² = 1; b² = 9

a = 1; b = 3

c² = a² + b²

= 1 + 9

= 10

c = \(\sqrt {10}\)

ae = \(\sqrt {10}\)

e = \(\sqrt {10}\)

centre (h, k) = (2, -3)

vertices (h ± a, k) = (2 ± 1, -3)

= (2 + 1, -3) & (2 – 1, -3)

= (3, -3) & (1, -3)

foci (h ± c, k) = (2 ±\(\sqrt {10}\), -3)

= (2 + \(\sqrt {10}\), -3) & (2 –\(\sqrt {10}\), -3)

directrix x = ±\(\frac {a}{e}\) + h

= ±\(\frac {1}{\sqrt {10}}\) + 2 & x = –\(\frac {1}{\sqrt {10}}\) + 2