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New answer posted•a year ago

Exemplar Solution 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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution 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.

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution 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) $y^{2} = 8 (x + 5)$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution 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.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Find the distance from the origin to (6, 6, 7).

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} C o o r d i n a t e s o f t h e o r i g i n a r e (0, 0, 0) \\ ∴ Distance f r o m (0, 0, 0) t o (6, 6, 7) = \sqrt[]{{(6 - 0)}^{2} + {(6 - 0)}^{2} + {(7 - 0)}^{2}} \\ = \sqrt[]{36 + 36 + 49} = \sqrt[]{121} = 11 u n i t s . \\ H e n c e, t h e r e q u i r e d = 11 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the parabola $y^{2} = 4 a x$ passes through the point (3, 2), then the length of its latus rectum is

(a) $\frac{2}{3}$
(b) 2
(c) 4
(d) None of these

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$G i v e n p a r a b o l a i s y^{2} = 4 a x I f t h e p a r a b o l a i s passing$
$\begin{array}{l} t h r o u g h (3, 2) \\ t h e n {(2)}^{2} = 4 a \times 3 \\ \Rightarrow 4 = 12 a \Rightarrow a = \frac{1}{3} \\ N o w l e n g t h o f t h e l a t u s r e c t u m = 4 a = 4 \times \frac{1}{3} = \frac{4}{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

How far apart are the points (2, 0, 0) and (–3, 0, 0)?

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n points a r e (2, 0, 0), a n d (- 3, 0, 0) \\ ∴ Distance b e t w e e n t h e g i v e n = \sqrt[]{{(2 + 3)}^{2} + {(0 - 0)}^{2} + {(0 - 0)}^{2}} = \sqrt[]{25} = 5 \\ H e n c e, t h e r e q u i r e d = 5 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Let  $A, B, C$ be the feet of perpendiculars from a point  $P$ on the $x y$ -, $y z$ -, and $z x$ -planes, respectively. Find the coordinates of  $A, B, C$ in each of the following where the point  $P$ is:

i. (3, 4, 5),

ii. (–5, 3, 7)

iii. (4, –3, –5).

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e c o o r d i n a t e s o f A, B a n d C a r e \\ (i) A (3, 4, 0), B (0, 4, 5) a n d C (3, 0, 5) \\ (i i) A (- 5, 3, 0), B (0, 3, 7) a n d C (- 5, 0, 7) \\ (i i i) A (4, - 3, 0), B (0, - 3, - 5) a n d C (4, 0, - 5) \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

If the focus of a parabola is (0, -3) and its directrix is $y = 3$ , then its equation is

(a) $x^{2} = - 12 y$
(b) $x^{2} = 12 y$
(c) $y^{2} = - 12 x$
(d) $y^{2} = 12 x$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A c c o r d i n g t o t h e d e f i n i t i o n o f p a r a b o l a \\ \sqrt[]{{(x - 0)}^{2} + {(y + 3)}^{2}} = | \frac{y - 3}{\sqrt[]{{(0)}^{2} + {(1)}^{2}}} | \\ \Rightarrow \sqrt[]{x^{2} + y^{2} + 9 + 6 y} = | y - 3 | \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ x^{2} + y^{2} + 9 + 6 y = y^{2} + 9 - 6 y \\ \Rightarrow x^{2} + 9 + 6 y = 9 - 6 y \\ \Rightarrow x^{2} = - 12 y \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Let  $A, B, C$ be the feet of perpendiculars from a point  $P$ on the $x$ -, $y$ -, and $z$ -axes, respectively. Find the coordinates of  $A, B, C$ in each of the following where the point  $P$ is:

i. A = (3, 4, 2),

ii. (–5, 3, 7),

iii. (4, –3, –5).

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e c o o r d i n a t e s o f A, B a n d C a r e \\ (i) A (3, 0, 0), B (0, 4, 0) a n d C (0, 0, 2) \\ (i i) A (- 5, 0, 0), B (0, 3, 0) a n d C (0, 0, 7) \\ (i i i) A (4, 0, 0), B (0, - 3, 0) a n d C (0, 0, - 5) \end{array}$

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