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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 y2 = 4 ax
Here a = 4
⇒ y2 = 4 (4) x (i.e.,)y2 = 16x
12th Maths Exercise 5.2 Samacheer Kalvi Chapter 5 Two Dimensional Analytical Geometry

(ii) The parabola is symmetric about y axis. So the equation will be of the form
x2 = 4 ay
It passes through (2, -3)
⇒ 22 = 4a(-3)
4 = -12a ⇒ a = \(-\frac{1}{3}\) ⇒ 4a = \(-\frac{4}{3}\)
∴ Equation of parabola is x2 = \(-\frac{4}{3}\) y
3x2= – 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 y2 = 4ax
Here a = 3 ⇒ y2 = 12x
but the vertex is (1, -2)
So equation of the parabola is
(y + 2)2 = 12(x – 1)
12th Maths Exercise 5.2 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi

(iv) Focus = (4, 0)
Equation of the parabola will be of the form .
y2 = 4ax
Here a = 4
⇒ y2 = 16x
12th Maths Exercise 5.2 8th Sum Maths Solutions Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi

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 a2 = 36
b2 = a2(1 – e2) = 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.
12th Maths 5th Chapter Samacheer Kalvi Two Dimensional Analytical Geometry - Ii Ex 5.2
Also a = 5 and ae = 4
⇒ 5e = 4 ⇒ e = \(\frac{4}{5}\)
Now a = 5 ⇒ a2 = 25
ae = 4 ⇒ ae2 = 16
We know b2 = a2 (1 – e2) = a2 – a2e2 = 25 – 16 = 9
Equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{25}\) = 1

12th Maths 5.2 Exercise Maths Solutions Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi

(iv) Given \(\frac{2 b^{2}}{a}\) = 4 and 2ae = \(4 \sqrt{2}\)
Now \(\frac{2 b^{2}}{a}\) = 4 2b2 = 4a
⇒ b2 = 2a
2ae = \(4 \sqrt{2}\) ae = \(2 \sqrt{2}\)
So a2e2 = 4(2) = 8
We know b2 = a2(1 – e2) = a2 – a2e2
⇒ 2a = a2 – 8 ⇒ a2 – 2a -8 = 0
⇒ (a – 4) (a +2) = 0 ⇒a = 4 or -2
As a cannot be negative
a = 4 So a2 = 16 and b2 = 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 a2 = \(\frac{16}{9}\)
b2 = a2(e2 – 1) = a2 e2 – a2 = 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 ⇒ a2 e2 = 36
(i.e.) (2e)2 (e)2 = 36
⇒ 4e4 = 36 ⇒ e4 = 9
⇒ e = \(\sqrt{3}\)
Now e = \(\sqrt{3}\) a = \(2\sqrt{3}\)
∴ a2 = 4 × 3 = 12
b2 = a2 (e2 – 1) = a2 e2 – a2 = 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
Class 12 Maths Chapter 5 Exercise 5.2 Two Dimensional Analytical Geometry Samacheer Kalvi

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) y2 = 16x
(ii) x2 = 24y
(iii) y2 = -8x
(iv) x2 – 2x + 8y + 17 = 0
(v) y2 – 4y – 8x + 12 = 0
Solution:
(i) y2 = 16x
It is of the form y2 = 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) x2 = 24y
This is of the form x2 = 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) y2 = -8x
This is of the form y2 = – 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) x2 – 2x + 8y + 17 = 0
x2 – 2x = -8y – 17
x2 – 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 X2 = – 8Y.
This is of the form x2 = – 4ay (type IV)
Where 4a = 8 ⇒ a = 2
12th Maths 5.2 Solutions Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi

(v) y2 – 4y = 8x – 12 = 0
y2 – 4y + 4 = 8x – 12 + 4
(y – 2)2 = 8x – 8 = 8 (x – 1)
Taking x – 1 = X and y – 2 = Y.
We get Y2 = 8X.
This is of the form y2 = 4ax (type IV)
Where 4a = 8 ⇒ a = 2
12th Maths Chapter 5 Exercise 5.2 Two Dimensional Analytical Geometry Samacheer Kalvi

12th Maths 5.2 Exercise Question 5.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:
Ex 5.2 Class 12  Maths Solutions Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi
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 a2 = 25, b2 = 9
a = 5, b = 3
e2 = \(\frac{a^{2}-b^{2}}{a^{2}}=\frac{25-9}{25}=\frac{16}{25}\) ⇒ 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)
10th Maths Exercise 5.2 Samacheer Kalvi Solutions Chapter 5 Two Dimensional Analytical Geometry
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
Samacheerkalvi.Guru 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - Ii Ex 5.2
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
5.2 Exercise Class 12 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi
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 Math 5.2 Solution Chapter 5 Two Dimensional Analytical Geometry Samacheer Kalvi

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:
12th Maths Book Samacheer Kalvi Chapter 5 Two Dimensional Analytical Geometry - Ii Ex 5.2
∴ 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:
12th Maths Samacheer Kalvi Solutions Chapter 5 Two Dimensional Analytical Geometry - Ii Ex 5.2
Solution:
Class 12 Ex5.2 Samacheer Kalvi Maths Solutions Chapter 5 Two Dimensional Analytical Geometry
Maths Class 12 Ex 5.2 Samacheer Kalvi Solutions Chapter 5 Two Dimensional Analytical Geometry
Ex5.2 Class 12 Samacheer Kalvi Maths Solutions Chapter 5 Two Dimensional Analytical Geometry
Exercise 5.2 Maths Class 12 Samacheer Kalvi Solutions Chapter 5 Two Dimensional Analytical Geometry
Exercise 5.2 Class 12 Maths Samacheer Kalvi Chapter 5 Two Dimensional Analytical Geometry
5.2 Class 12 Samacheer Kalvi Maths Solutions Chapter 5 Two Dimensional Analytical Geometry
Ch 5 Class 12 Maths Samacheer Kalvi Two Dimensional Analytical Geometry - Ii Ex 5.2

(v) 18x2 + 12y2 – 144x + 48y + 120 = 0
(18x2 – 144x) + (12y2 + 48y) = -120
18(x2 – 8x) + 12 (y2 + 4y) = -120
18(x2 – 8x + 16 – 16) + 12(y2 + 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 23
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 24
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 25

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:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 1

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:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 2
Here, centre = (1, -2) and transverse axis is parallel to X-axis.
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 3

12th Maths Chapter 5 Exercise 5.2 Question 3.
Find axis, vertex, focus and equation of directrix for y2 + 8x – 6y + 1 = 0
Solution:
y2 – 6y = – 8x – 1
y2 – 6y + 9 = – 8x – 1 + 9
(y – 3)2 = – 8x + 8 = – 8(x – 1)
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 4
Comparing this equation with Y2 = – 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 x2 – 6x – 12y – 3 = 0.
Solution:
x2 – 6x = 12y + 3
x2 – 6x + 9 = 12y + 3 + 9 = 12y + 12
(x – 3)2 = 12(y + 1)
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 40
This equation is of the form X2 = 4aY
4a = 12
⇒ a = 3
Vertex is x – 3 = 0 ; y + 1 = 0
⇒ x = 3 ; y = -1
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 41

Question 5.
Find the eccentricity, centre, foci and vertices of the following hyperbolas: x2 – 4y2 – 8x – 6y – 18 = 0
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 42
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 43

Question 6.
Find the eccentricity, centre, foci, vertices of 9x2 + 4y2 = 36
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 44

Ex 5.2 Class 12 Question 7.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
x2 – 4y2 + 6x + 16y – 11 = 0
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 15
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 16

10th Maths Exercise 5.2 Samacheer Kalvi Question 8.
Find the eccentricity, centre, foci and vertices of the following hyperbolas:
x2 – 3y2 + 6x + 6y + 18 = 0
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 17

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