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

**12th Maths Exercise 5.2 Samacheer Kalvi 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 = F = (4, 0)

⇒ a = 4

Equation of directrix x = – 4

⇒ The curve open to the right. So the equation will be of the form y^{2} = 4 ax

Here a = 4

⇒ y^{2} = 4 (4) x (i.e.,)y^{2} = 16x

(ii) The parabola is symmetric about y axis. So the equation will be of the form

x^{2} = 4 ay

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) The distance between vertex and focus = 3

(ie.,) a = 3

Parabola is open to the right.

So equation will be of the form y^{2} = 4ax

Here a = 3 ⇒ y^{2} = 12x

but the vertex is (1, -2)

So equation of the parabola is

(y + 2)^{2} = 12(x – 1)

(iv) Focus = (4, 0)

Equation of the parabola will be of the form .

y^{2} = 4ax

Here a = 4

⇒ y^{2} = 16x

**12th Maths Exercise 5.2 Question 2.**

ind 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 \(4 \sqrt{2}\) and major axis as y -axis.

Solution:

(i) Given ae = 3 and e = \(\frac{1}{2}\)

⇒ a(\(\frac{1}{2}\)) = 3 ⇒ a = 6

So a^{2} = 36

b^{2} = a^{2}(1 – e^{2}) = 36 (1 – \(\frac{1}{4}\)) = 36 × \(\frac{3}{4}\) = 27

Since Foci = (± 3, 0), major axis is along x-axis

So equation of ellipse is \(\frac{x^{2}}{36}+\frac{y^{2}}{27}\) = 1

(ii) From the diagram we see that major axis is along y-axis.

Also a = 5 and ae = 4

⇒ 5e = 4 ⇒ e = \(\frac{4}{5}\)

Now a = 5 ⇒ a^{2} = 25

ae = 4 ⇒ ae^{2} = 16

We know b^{2} = a^{2} (1 – e^{2}) = a^{2} – a^{2}e^{2} = 25 – 16 = 9

Equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{25}\) = 1

(iv) 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

**12th Maths Exercise 5.2 8th Sum 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 JC axis and of length 8 units.

Solution:

(i) Given

ae = 2 and e = \(\frac{3}{2}\)

a( \(\frac{3}{2}\)) = 2 ⇒ a = \(\frac{4}{3}\) So a^{2} = \(\frac{16}{9}\)

b^{2} = a^{2}(e^{2} – 1) = a^{2} e^{2} – a^{2} = 4 – \(\frac{16}{9}\) = \(\frac{20}{9}\)

Since the foci are (± 2, 0), transverse axis is along x-axis

So equation of hyperbola is

\(\frac{x^{2}}{16 / 9}-\frac{y^{2}}{20 / 9}\) = 1 ⇒ \(\frac{9 x^{2}}{16}-\frac{9 y^{2}}{20}\) = 1

(ii) Given Centre = (2, 1)

ae = 6 (distance between (2, 1) and (8, 1)) ……………. (1)

Also \(\frac{a}{e}\) = 2 ⇒ a = 2e

Equation of directrix is x = 4 [(i.e.,) (x – 2 = 2) Since centre is (2, 1)]

⇒ \(\frac{a}{e}\) = 2

Given ae = 6 ⇒ a^{2} e^{2} = 36

(i.e.) (2e)^{2} (e)^{2} = 36

⇒ 4e^{4} = 36 ⇒ e^{4} = 9

⇒ e = \(\sqrt{3}\)

Now e = \(\sqrt{3}\) a = \(2\sqrt{3}\)

∴ a^{2} = 4 × 3 = 12

b^{2} = a^{2} (e^{2} – 1) = a^{2} e^{2} – a^{2} = 36 – 12 = 24

So here Centre = (2, 1)

So equation of hyperbola is

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

(iii) Length of the transverse axis = 8

2a = 8 ⇒ a = 4

Transverse axis is along x-axis

So of equation of hyperbola is will be

**12th Maths 5th Chapter Samacheer Kalvi Question 4.**

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

(i) y^{2} = 16x

(ii) x^{2} = 24y

(iii) y^{2} = -8x

(iv) x^{2} – 2x + 8y + 17 = 0

(v) y^{2} – 4y – 8x + 12 = 0

Solution:

(i) y^{2} = 16x

It is of the form y^{2} = 4ax (type I)

Here 4a = 16 ⇒ a = 4

Vertex = (0, 0)

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

Equation of directrix x + 4 = 0 (or) x = – 4

Length of latus rectum = 4a = 16 .

(ii) x^{2} = 24y

This is of the form x^{2} = 4ay (type III)

4a = 24 ⇒ a = 6

Vertex = (0, 0)

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

Equation of directrix is y + a= 0 (i.e.,) y + 6 = 0 (or) y = -6

Length of latus rectum = 4a = 24.

(iii) y^{2} = -8x

This is of the form y^{2} = – 4ax (type II)

Here 4a = 8 ⇒ a = 2

Vertex = (0, 0)

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

Equation of directrix is x – 2 = 0 (or) x = 2

Length of latus rectum = 4a = 8.

(iv) x^{2} – 2x + 8y + 17 = 0

x^{2} – 2x = -8y – 17

x^{2} – 2x + 1 – 1 = – 8y — 17

(x – 1)^{2} = – 8y – 17 + 1 = – 8y + 16

(x – 1)^{2} = – 8 (y – 2)

Taking x – 1 = X and y – 2 = Y.

We get X^{2} = – 8Y.

This is of the form x^{2} = – 4ay (type IV)

Where 4a = 8 ⇒ a = 2

(v) y^{2} – 4y = 8x – 12 = 0

y^{2} – 4y + 4 = 8x – 12 + 4

(y – 2)^{2} = 8x – 8 = 8 (x – 1)

Taking x – 1 = X and y – 2 = Y.

We get Y^{2} = 8X.

This is of the form y^{2} = 4ax (type IV)

Where 4a = 8 ⇒ a = 2

**12th Maths 5.2 Exercise Question 5.**

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

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 and here (always a >b)

Here the major axis is alongy-axis

Centre = (0, 0)

Foci = (0, ± ae) = (0, ± \(\sqrt{7}\))

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

Equation of directrices y = ± \(\frac{10}{\sqrt{7}}\)

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

Now, Here transverse axis is along x-axis

Centre = (0, 0)

Vertices = (± a, 0) = (± 5, 0)

Foci = (± ae, 0) = (± 13, 0)

Equation of directrices x = ± \(\frac{a}{e}\) (ie.,) x = ± \(\frac{25}{13}\)

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

It is a hyperbola. Here transverse axis is along y-axis

Now Centre = (0, 0)

Vertices = (0, ± a) = (0, ± 4)

Foci = (0, ± ae) = (0, ± 5)

Equation of directrices y = ± \(\frac{16}{5}\)

**Class 12 Maths Chapter 5 Exercise 5.2 Question 6.**

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

Solution:

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

**12th Maths 5.2 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:

∴ S’P – SP = (a + ex) – (ex – a)

a + ex – ex + a = 2a (transverse axis)

Question 8.

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

Solution:

(v) 18x^{2} + 12y^{2} – 144x + 48y + 120 = 0

(18x^{2} – 144x) + (12y^{2} + 48y) = -120

18(x^{2} – 8x) + 12 (y^{2} + 4y) = -120

18(x^{2} – 8x + 16 – 16) + 12(y^{2} + 4y + 4 – 4) = -120

18(x – 4)^{2} – 288 + 12(y + 2)^{2} – 48 = – 120

18(x – 4)^{2} + 12(y + 2)^{2} = -120 + 288 + 48 = 216

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

Question 1.

Find the equation of the ellipse if centre is (3, – 4), one of the foci is (3 + \(\sqrt{3}\), – 4) and e = \(\frac{\sqrt{3}}{2}\)

Solution:

Question 2.

Find the equation of the hyperbola if centre (1, -2); length of the transverse axis is 8; e = \(\frac{5}{4}\) and the transverse axis is parallel to X-axis.

Solution:

Here, centre = (1, -2) and transverse axis is parallel to X-axis.

**12th Maths Chapter 5 Exercise 5.2 Question 3.**

Find axis, vertex, focus and equation of directrix for y^{2} + 8x – 6y + 1 = 0

Solution:

y^{2} – 6y = – 8x – 1

y^{2} – 6y + 9 = – 8x – 1 + 9

(y – 3)^{2} = – 8x + 8 = – 8(x – 1)

Comparing this equation with Y^{2} = – 4aX we get

4a = 8 or a = 2

Vertex is (0, 0)

x – 1 = 0 ⇒ x = 1, y – 3 = 0 ⇒ y = 3

Axis y – 3 = 0, Vertex = (1, 3)

Focus is (- a, 0) = (-2, 0) + (1, 3) = (-1, 3)

Equation of directrix is x – a = 0. i.e., X – 2 = 0

⇒ x – 1 – 2 = 0 ⇒ x – 3 = 0

Latus rectum x + a = 0 i.e., x – 1 + 2 = 0

x + 1 = 0

Length of latus rectum = 4a = 8

Question 4.

Find axis, Vertex focus and equation of directrix for x^{2} – 6x – 12y – 3 = 0.

Solution:

x^{2} – 6x = 12y + 3

x^{2} – 6x + 9 = 12y + 3 + 9 = 12y + 12

(x – 3)^{2} = 12(y + 1)

This equation is of the form X^{2} = 4aY

4a = 12

⇒ a = 3

Vertex is x – 3 = 0 ; y + 1 = 0

⇒ x = 3 ; y = -1

Question 5.

Find the eccentricity, centre, foci and vertices of the following hyperbolas: x^{2} – 4y^{2} – 8x – 6y – 18 = 0

Solution:

Question 6.

Find the eccentricity, centre, foci, vertices of 9x^{2} + 4y^{2} = 36

Solution:

**Ex 5.2 Class 12 Question 7.**

Find the eccentricity, centre, foci and vertices of the following hyperbolas:

x^{2} – 4y^{2} + 6x + 16y – 11 = 0

Solution:

**10th Maths Exercise 5.2 Samacheer Kalvi Question 8.**

Find the eccentricity, centre, foci and vertices of the following hyperbolas:

x^{2} – 3y^{2} + 6x + 6y + 18 = 0

Solution: