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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

The mid-points of the sides of a triangle are <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>5</mn> <mo>,</mo> <mn>7</mn> <mo>,</mo> <mn>1</mn> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> , <math> <mrow> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <mn>8</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> , and <math> <mrow> <mrow> <mo>(</mo> <mrow><mn>2</mn><mo>,</mo><mn>3</mn><mo>,</mo>–<mn>1</mn></mrow> <mo>)</mo> </mrow> </mrow> </math> . Find its vertices.

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Asked by Payal Gupta

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the hyperbola with eccentricity $\sqrt[]{2}$ and foci at $(\pm 2, 0)$ is

(a) $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
(c) $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
(d) None of these

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

^{$\begin{array}{l} G i v e n t h a t e = \frac{3}{2} \\ a n d f o c i = (\pm a e, 0) = (\pm 2, 0) \\ ∴ a e = 2 \Rightarrow a \times \frac{3}{2} = 2 \Rightarrow a = \frac{4}{3} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = \frac{16}{9} (\frac{9}{4} - 1) \Rightarrow b^{2} = \frac{16}{9} \times \frac{5}{4} \\ \Rightarrow b^{2} = \frac{20}{9} \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{{(4 / 3)}^{2}} - \frac{y^{2}}{20 / 9} = 1 \Rightarrow \frac{9 x^{2}}{16} - \frac{9 y^{2}}{20} = 1 \Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{20} = \frac{1}{9} \\ \Rightarrow \frac{x^{2}}{4} - \frac{y^{2}}{5} = \frac{4}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$}

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The distance between the foci of a hyperbola is 16, and its eccentricity is 2. Its equation is

(a) $x^{2} - y^{2} = 32$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$
(c) $2 x - 3 y^{2} = 7$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$W e k n o w t h a t Distance$
$\begin{array}{l} b e t w e e n t h e f o c i = 2 a e \\ ∴ 2 a e = 16 \Rightarrow a e = 8 \\ G i v e n t h a t e = \sqrt[]{2} \\ ∴ \sqrt[]{2} a = 8 \Rightarrow a = 4 \sqrt[]{2} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 32 (2 - 1) \Rightarrow b^{2} = 32 \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x^{2}}{32} - \frac{y^{2}}{32} = 1 \Rightarrow x^{2} - y^{2} = 32 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Find the centroid of a triangle, the midpoints of whose sides are $D (1, 2, – 3)$ , $E (3, 0, 1)$ , and $F (– 1, 1, – 4)$ .

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e v e r t i c e s o f Δ A B C b e \\ A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}) a n d C (x_{3}, y_{3}, z_{3}) \\ M i d point o f B C = (1, 2, - 3) \\ ∴ 1 = \frac{x_{2} + x_{3}}{2} \Rightarrow x_{2} + x_{3} = 2 \dots (i) \\ 2 = \frac{y_{2} + y_{3}}{2} \Rightarrow y_{2} + y_{3} = 4 \dots (i i) \\ a n d - 3 = \frac{z_{2} + z_{3}}{2} \Rightarrow z_{2} + z_{3} = - 6 \dots (i i i) \\ M i d point o f A B = (3, 0, 1) \\ ∴ 3 = \frac{x_{1} + x_{2}}{2} \Rightarrow x_{1} + x_{2} = 6 \dots (i v) \\ 0 = \frac{y_{1} + y_{2}}{2} \Rightarrow y_{1} + y_{2} = 0 \dots (v) \\ a n d 1 = \frac{z_{1} + z_{2}}{2} \Rightarrow z_{1} + z_{2} = 2 \dots (v i) \\ S i m i l a r l y, M i d point o f A C = (- 1, 1, - 4) \\ ∴ - 1 = \frac{x_{1} + x_{3}}{2} \Rightarrow x_{1} + x_{3} = - 2 \dots (v i i) \\ 1 = \frac{y_{1} + y_{3}}{2} \Rightarrow y_{1} + y_{3} = 2 \dots (v i i i) \\ a n d - 4 = \frac{z_{1} + z_{3}}{2} \Rightarrow z_{1} + z_{3} = - 8 \dots (i x) \\ A d d i n g e q . (i), (i v) a n d (v i i) w e g e t, \\ 2 x_{1} + 2 x_{2} + 2 x_{3} = 2 + 6 - 2 = 6 \\ \Rightarrow x_{1} + x_{2} + x_{3} = 3 \\ \Rightarrow 6 + x_{3} = 3 \Rightarrow x_{3} = - 3 [? F r o m e q . (i v)] \\ \Rightarrow x_{1} + 2 = 3 \Rightarrow x_{1} = 1 [? F r o m e q . (i)] \\ \Rightarrow x_{2} - 2 = 3 \Rightarrow x_{2} = 5 [? F r o m e q . (v i i)] \\ S o, x_{1} = 1, x_{2} = 5 a n d x_{3} = - 3 \\ S i m i l a r l y, A d d i n g e q . (i i), (v) a n d (v i i i) w e g e t, \\ 2 (y_{1} + y_{2} + y_{3}) = 4 + 0 + 2 = 6 \\ \Rightarrow y_{1} + y_{2} + y_{3} = 3 \end{array}$

$\begin{array}{l} \Rightarrow y_{1} + 4 = 3 \Rightarrow y_{1} = - 1 \\ \Rightarrow 0 + y_{3} = 3 \Rightarrow y_{3} = 3 \\ \Rightarrow y_{2} + 2 = 3 \Rightarrow y_{2} = 1 \\ S o, y_{1} = - 1, y_{2} = 1 a n d y_{3} = 3 \\ A d d i n g e q . (i i i), (v i) a n d (i x) w e g e t, \\ 2 (z_{1} + z_{2} + z_{3}) = - 6 + 2 - 8 = - 12 \\ \Rightarrow z_{1} + z_{2} + z_{3} = - 6 \\ \Rightarrow z_{1} - 6 = - 6 \Rightarrow z_{1} = 0 [? F r o m e q . (i i i)] \\ \Rightarrow 2 + z_{3} = - 6 \Rightarrow z_{3} = - 8 [? F r o m e q . (v i)] \\ \Rightarrow z_{2} - 8 = - 6 \Rightarrow z_{2} = 2 \\ S o, z_{1} = 1, z_{2} = 1 a n d z_{3} = - 8 . \\ S o, t h e points a r e A (1, - 1, 0), B (5, 1, 2) a n d C (- 3, 3, - 8) . \\ ∴ C e n t r o i d o f t h e t r i a n g l e \\ G = (\frac{1 + 5 - 3}{3}, \frac{- 1 + 1 + 3}{3}, \frac{0 + 2 - 8}{3}) = (1, 1, - 2) \\ H e n c e, t h e r e q u i r e d c< & < \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is half the distance between the foci is (a) $\frac{3}{2}$
(b) $2$
(c) $\frac{3}{2}$
(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e n g t h o f t h e l a t u s r e c t u m o f t h e h y p e r b o l a \\ = \frac{2 b^{2}}{a} = 8 \Rightarrow b^{2} = 4 a \dots (i) \\ Distance b e t w e e n t h e f o c i = 2 a e \\ T r a n s v e r s e a x i s = 2 a \\ a n d C o n j u g a t e a x i s = 2 b \\ ∴ \frac{1}{2} (2 a e) = 2 b \Rightarrow a e = 2 b \Rightarrow b = \frac{a e}{2} \dots (i i) \\ \Rightarrow b^{2} = \frac{a^{2} e^{2}}{4} \Rightarrow 4 a = \frac{a^{2} e^{2}}{4} [F r o m e q n . (i)] \\ \Rightarrow 16 = a e^{2} ∴ a = \frac{16}{e^{2}} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow 4 a = a^{2} (e^{2} - 1) \\ \Rightarrow \frac{4}{a} = e^{2} - 1 \Rightarrow \frac{4}{16 / e^{2}} = e^{2} - 1 \\ \Rightarrow \frac{e^{2}}{4} = e^{2} - 1 \Rightarrow e^{2} - \frac{e^{2}}{4} = 1 \Rightarrow \frac{3 e^{2}}{4} = 1 \\ \Rightarrow e^{2} = \frac{4}{3} ∴ e = \frac{2}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Find the third vertex of a triangle whose centroid is the origin and two vertices are $(2, 4, 6)$  and $(0, – 2, – 5)$ .

6 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e t h i r d v e r t e x i . e ., A b e (a, b, c) \\ Since t h e c e n t r o i d i s a t o r i g i n i . e, (0, 0, 0) \\ ∴ 0 = \frac{a + 2 + 0}{2} \Rightarrow a = - 2 \\ 0 = \frac{b + 4 - 2}{2} \Rightarrow b = - 2 \\ a n d 0 = \frac{c + 6 - 5}{2} \Rightarrow c = - 1 \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (- 2, - 2, - 1) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If $e$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (where $a < b$ ), then

(a) $b^{2} = a^{2} (1 - e^{2})$
(b) $a^{2} = b^{2} (1 - e^{2})$
(c) $b^{2} = a^{2} (e^{2} - 1)$
(d) $a^{2} = b^{2} (e^{2} - 1)$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b) \\ ∴ E c c e n t r i c i t y e = \sqrt[]{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e^{2} = 1 - \frac{a^{2}}{b^{2}} \\ \Rightarrow \frac{a^{2}}{b^{2}} = (1 - e^{2}) \Rightarrow a^{2} = b^{2} (1 - e^{2}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Show that the triangle  $A B C$ with vertices $A (0, 4, 1)$ , $B (2, 3, – 1)$ , $C (4, 5, 0)$ and  is right angled.

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n v e r t i c e s a r e A (0, 4, 1), B (2, 3, - 1) a n d C (4, 5, 0) \\ A B = \sqrt[]{{(2 - 0)}^{2} + {(3 - 4)}^{2} + {(- 1 - 1)}^{2}} = \sqrt[]{4 + 1 + 4} = \sqrt[]{9} = 3 \\ B C = \sqrt[]{{(4 - 2)}^{2} + {(5 - 3)}^{2} + {(0 + 1)}^{2}} = \sqrt[]{4 + 4 + 1} = \sqrt[]{9} = 3 \\ A C = \sqrt[]{{(4 - 0)}^{2} + {(5 - 4)}^{2} + {(0 - 1)}^{2}} = \sqrt[]{16 + 1 + 1} = \sqrt[]{18} \\ ? {(3)}^{2} + {(3)}^{2} = {(\sqrt[]{18})}^{2} \\ S o, A B^{2} + B C^{2} = A C^{2} . \\ H e n c e, Δ A B C i s a r i g h t a n g l e d t r i a n g l e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The length of the latus rectum of the ellipse $3 x^{2} + y^{2} = 12$ is

(a) 4
(b) 3
(c) $\frac{8}{3}$
(d) $\frac{4}{3}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f t h e e l l i p s e i s 3 x^{2} + y^{2} = 12 \\ \Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{12} = 1 \\ H e r e a^{2} = 4 \Rightarrow a = 2 \\ b^{2} = 12 \Rightarrow b = 2 \sqrt[]{3} \\ L e n g t h o f t h e l a t u s r e c t u m = \frac{2 a^{2}}{b} = \frac{2 \times 4}{2 \sqrt[]{3}} = \frac{4}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Three consecutive vertices of a parallelogram $A B C D$  are $A (6, – 2, 4)$ , $B (2, 4, – 8)$ , and $C (– 2, 2, 4)$ . Find the coordinates of the fourth vertex.

[Hint: Diagonals of a parallelogram have the same mid-point.]

10 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e f o u r t h v e r t e x b e (a, b, c) \\ W e k n o w t h a t t h e d i a g o n a l s o f a parallelogram bisect e a c h o t h e r . \\ ∴ M i d p o i n t o f d i a g o n a l A C = (\frac{6 - 2}{2}, \frac{- 2 + 2}{2}, \frac{4 + 4}{2}) = (2, 0, 4) \\ a n d t h e m i d p o i n t o f d i a g o n a l B D = (\frac{a + 2}{2}, \frac{b + 4}{2}, \frac{c - 8}{2}) \\ ∴ \frac{a + 2}{2} = 2 \Rightarrow a = 2 \\ \frac{b + 4}{2} = 0 \Rightarrow b = - 4 \\ \frac{c - 8}{2} = 4 \Rightarrow c = 16 \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (2, - 4, 16) . \end{array}$

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The equation of the hyperbola with eccentricity $\sqrt[]{2}$ and foci at $(\pm 2, 0)$ is

(a) $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
(c) $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
(d) None of these

The distance between the foci of a hyperbola is 16, and its eccentricity is 2. Its equation is

(a) $x^{2} - y^{2} = 32$
(b) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$
(c) $2 x - 3 y^{2} = 7$
(d) None of these

Find the centroid of a triangle, the midpoints of whose sides are $D (1, 2, – 3)$ , $E (3, 0, 1)$ , and $F (– 1, 1, – 4)$ .

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is half the distance between the foci is (a) $\frac{3}{2}$
(b) $2$
(c) $\frac{3}{2}$
(d) None of these

Find the third vertex of a triangle whose centroid is the origin and two vertices are $(2, 4, 6)$  and $(0, – 2, – 5)$ .

If $e$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (where $a < b$ ), then

(a) $b^{2} = a^{2} (1 - e^{2})$
(b) $a^{2} = b^{2} (1 - e^{2})$
(c) $b^{2} = a^{2} (e^{2} - 1)$
(d) $a^{2} = b^{2} (e^{2} - 1)$

Show that the triangle  $A B C$ with vertices $A (0, 4, 1)$ , $B (2, 3, – 1)$ , $C (4, 5, 0)$ and  is right angled.

The length of the latus rectum of the ellipse $3 x^{2} + y^{2} = 12$ is

(a) 4
(b) 3
(c) $\frac{8}{3}$
(d) $\frac{4}{3}$