Tags Maths

Maths

Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths

Follow Ask Question

6.5k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Maths 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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

Maths 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 $(0, – 2, – 5)$ .

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

Maths 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.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

Maths 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) $\frac{4}{3}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

Maths 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.]

0 Follower 8 Views

Payal Gupta

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the ellipse whose focus is (1, â€“1), the directrix is the line $x - y - 3 = 0$ , and the eccentricity is $\frac{1}{2}$ is

(a) $7 x^{2} + 2 x y + 7 y^{2} - 10 x + 10 y + 7 = 0$
(b) $7 x^{2} + 2 x y + 7 y^{2} + 7 = 0$
(c) $7 x^{2} + 2 x y + 7 y^{2} + 10 x - 10 y - 7 = 0$
(d) None of these

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f o c u s o f t h e e l l i p s e i s (1, - 1) a n d t h e e q u a t i o n o f t h e d i r e c t r i x i s x - y - 3 = 0 \\ a n d e = \frac{1}{2} . \\ L e t P (x, y) b e a n y point o n t h e p a r a b o l a \\ ∴ \frac{P F}{Distance o f t h e point P f r o m t h e d i r e c t r i x} = e \\ \Rightarrow \frac{\sqrt[]{{(x - 1)}^{2} + {(y + 1)}^{2}}}{| \frac{x - y - 3}{\sqrt[]{{(1)}^{2} + {(- 1)}^{2}}} |} = \frac{1}{2} \\ \Rightarrow 2 \sqrt[]{x^{2} + 1 - 2 x + y^{2} + 1 + 2 y} = | \frac{x - y - 3}{\sqrt[]{2}} | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ \Rightarrow 4 (x^{2} + y^{2} - 2 x + 2 y + 2) = \frac{x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x}{2} \\ \Rightarrow 8 x^{2} + 8 y^{2} - 16 x + 16 y + 16 = x^{2} + y^{2} + 9 - 2 x y + 6 y - 6 x \\ \Rightarrow 7 x^{2} + 7 y^{2} + 2 x y - 10 x + 10 y + 7 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Show that the points $A (1, – 1, 3)$ , $B (2, – 4, 5)$ , and $(5, – 13, 11)$ are collinear.

0 Follower 10 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n points a r e A (1, - 1, 3), B (2, - 4, 5) a n d C (5, - 13, 11) \\ A B = \sqrt[]{{(2 - 1)}^{2} + {(- 4 + 1)}^{2} + {(5 - 3)}^{2}} = \sqrt[]{1 + 9 + 4} = \sqrt[]{14} \\ B C = \sqrt[]{{(5 - 2)}^{2} + {(- 13 + 4)}^{2} + {(11 - 5)}^{2}} = \sqrt[]{9 + 81 + 36} = \sqrt[]{126} = 3 \sqrt[]{14} \\ A C = \sqrt[]{{(5 - 1)}^{2} + {(- 13 + 1)}^{2} + {(11 - 3)}^{2}} = \sqrt[]{16 + 144 + 64} = \sqrt[]{224} = 4 \sqrt[]{14} \\ H e r e, w e o b s e r v e t h a t \sqrt[]{14} + 3 \sqrt[]{14} = 4 \sqrt[]{14} \\ S o, A B + B C = A C . \\ H e n c e, t h e g i v e n points a r e c o l l i n e a r . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the vertex of the parabola is at the point (â€“3, 0) and the directrix is the line $x + 5 = 0$ , then its equation is

(a) $y^{2} = - 8 (x + 3)$
(b) $x^{2} = 8 (y + 3)$
(c) $y^{2} = 8 (x + 3)$
(d) $y^{2} = 8 (x + 5)$

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t v e r t e x = (- 3, 0) \\ ∴ a = - 3 a n d d i r e c t r i x i s x + 5 = 0 \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a, w e g e t \\ A F = A D i . e ., A i s t h e m i d point o f D F \\ ∴ - 3 = \frac{x_{1} - 5}{2} \Rightarrow x_{1} = - 6 + 5 = - 1 \\ a n d 0 = \frac{0 + y_{1}}{2} \Rightarrow y_{1} = 0 \\ ∴ F o c u s F = (- 1, 0) \\ N o w \sqrt[]{{(x + 1)}^{2} + {(y - 0)}^{2}} = | \frac{x + 5}{\sqrt[]{1^{2} + 0^{2}}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ {(x + 1)}^{2} + {(y - 0)}^{2} = {(x + 5)}^{2} \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 25 + 10 x \\ \Rightarrow y^{2} = 10 x - 2 x + 24 \Rightarrow y^{2} = 8 x + 24 \\ \Rightarrow y^{2} = 8 (x + 3) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Show that if $x^{2} + y^{2} = 1$ , then the point $(x, y, \sqrt[]{1 - x^{2} - y^{2}})$ is at a distance of 1 unit from the origin.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n i s (x, y, \sqrt[]{1 - x^{2} - y^{2}}) \\ ∴ Distance b e t w e e n t h e o r i g i n a n d t h e point i s = \sqrt[]{{(x - 0)}^{2} + {(y - 0)}^{2} + {(\sqrt[]{1 - x^{2} - y^{2}} - 0)}^{2}} \\ = \sqrt[]{1} = 1 \\ H e n c e p r o v e d . \end{array}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
711k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Maths

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience