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

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

(ii) The parabola is symmetric about y axis. So the equation will be of the form
In given data the parabola is open downwards, vertex be (0, 0)
Equation of parabola x² = -4ay
It passes through point (2, -3)
4 = -4a (-3)
1 = 3a
a = \(\frac {1}{3}\)
∴ Equation of parabola
x² = -4(\(\frac {1}{3}\)) 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)
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 2

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

(iv) Equation of parabola
y² = 4ax
a = VS = 4
y² = 4(4)x
y² = 16x
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 3

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 \(4 \sqrt{2}\) 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.
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 4
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

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

(iv) Distance between foci = 4√2
2ae = 4√2
ae = 2√2
Length of latus rectum = 4
\(\frac {2b^2}{a}\) = 4
b² = 2a
b² = a²(1 – e²)
2a = a² – a² e²
2a = a² – (2√2)²
⇒ a² – 8 – 2a = 0
(a – 4) (a + 2) = 0
a = 4 & a = -2 (Not possible)
∴ a = 4 ⇒ a² = 16
b² = 2a = 2(4) = 8
\(\frac {x^2}{b^2}\) + \(\frac {y^2}{a^2}\) = 1
\(\frac {x^2}{8}\) + \(\frac {y^2}{16}\) = 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 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
The transverse axis is along the x-axis
So of the equation of the hyperbola is will be
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 6

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
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 7

(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
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 8

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

Question 5.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 9
Solution:
(i) \(\frac {x^2}{25}\) + \(\frac {y^2}{9}\) = 1
It is an ellipse and the major axis is along x-axis.
a² = 25; b² = 9
a = 5; b = 3
c² = a² – b²
= 25 – 9 = 16
c = 4
ae = 4
⇒ 5e = 4
e = \(\frac {4}{5}\)
Centre = (0, 0)
Vertex (±a, 0) = (±5, 0)
Foci (±c, 0) = (±4, 0)
Equation of directrix
x = ±\(\frac {a}{e}\) = ±\(\frac {5}{\frac{4}{5}}\) = ±\(\frac {25}{4}\)

(ii) \(\frac{x^{2}}{3}+\frac{y^{2}}{10}\) = 1
It is an ellipse and here (always a >b)
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 10
Here the major axis is along y-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
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 11
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
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 12
Now Centre = (0, 0)
Vertices = (0, ± a) = (0, ± 4)
Foci = (0, ± ae) = (0, ± 5)
Equation of directrices y = ± \(\frac{16}{5}\)

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)
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 13

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:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 14
∴ S’P – SP = (a + ex) – (ex – a)
a + ex – ex + a = 2a (transverse axis)

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

Question 8.
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 15
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 16
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 17
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 18
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 19
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 20
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 21
Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.2 22

(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

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

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

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

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

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

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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