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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of a circle with the origin as the center and passing through the vertices of an equilateral triangle whose median is of length $3 a$ is

$(a)$ (a) $x^{2} + y^{2} = 9 a^{2}$
(b) $x^{2} + y^{2} = 9 a^{2}$
(c) $x^{2} + y^{2} = 4 a^{2}$
(d) $x^{2} + y^{2} = a^{2}$

[Hint: The centroid of the triangle coincides with the center of the circle, and the radius is $\frac{2}{3}$ of the median.]

8 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:

$\begin{array}{l} L e t A B C b e a n e q u i l a t e r a l t r i a n g l e i n w h i c h m e d i a n A D = 3 a . \\ C e n t r e o f t h e c i r c l e i s s a m e a s t h e c e n t r o i d o f t h e t r i a n g l e i . e ., (0, 0) \\ A G : G D = 2 : 1 \\ S o, A G = \frac{2}{3} A D = \frac{2}{3} \times 3 a = 2 a \\ ∴ T h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 0)}^{2} + {(y - 0)}^{2} = {(2 a)}^{2} \\ \Rightarrow x^{2} + y^{2} = 4 a^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Name the octant in which each of the following points lies:

i. (1, 2, 3),

ii. (4, –2, 3),

iii. (4, –2, –5),

iv. (4, 2, –5),

v. (–4, 2, 5),

vi. (–3, –1, 6),

vii. (2, –4, –7),

viii. (–4, 2, –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

$\begin{array}{l} (i)Point (1, 2, 3) l i e s i n IOctant \\ (i i)Point (4, - 2, 3) l i e s i n I VOctant \\ (i i i)Point (4, - 2, - 5) l i e s i n V I I I Octant \\ (i v)Point (4, 2, - 5) l i e s i n VOctant \\ (v)Point (- 4, 2, 5) l i e s i n I IOctant \\ (v i)Point (- 3, - 1, 6) l i e s i n I I IOctant \\ (v i i)Point (2, - 4, - 7) l i e s i n V I I IOctant \\ (v i i i)Point (- 4, 2, - 5) l i e s i n V IOctant \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the circle with its center on the y-axis and passing through the origin and the point (2, 3) is

(a) $x^{2} + y^{2} + 13 y = 0$
(b) $x^{2} + y^{2} + 13 x + 3 = 0$
(c) $x^{2} + y^{2} - 13 y = 0$
(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:

$\begin{array}{l} L e t t h e e q u a t i o n o f t h e c i r c l e b e \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ L e t t h e c e n t r e b e (0, a) a n d r = a \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - 0)}^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} + {(y - a)}^{2} = a^{2} \\ \Rightarrow x^{2} + y^{2} + a^{2} - 2 a y = a^{2} \\ \Rightarrow x^{2} + y^{2} - 2 a y = 0 \dots (i) \\ N o w C P = r \\ \Rightarrow \sqrt[]{{(2 - 0)}^{2} + {(3 - a)}^{2}} = a \\ \Rightarrow \sqrt[]{4 + 9 + a^{2} - 6 a} = a \\ \Rightarrow \sqrt[]{13 + a^{2} - 6 a} = a \\ \Rightarrow 13 + a^{2} - 6 a = a^{2} \\ \Rightarrow 13 - 6 a = 0 \\ \Rightarrow a = \frac{13}{6} \\ P u t t i n g t h e v a l u e o f a i n e q n . (i) w e g e t \\ \Rightarrow x^{2} + y^{2} - 2 (\frac{13}{6}) y = 0 \\ \Rightarrow 3 x^{2} + 3 y^{2} - 13 y = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Locate the following points:

i. (1, –1, 3),

ii. (–1, 2, 4)

iii. (–2, –4, –7),

iv. (–4, 2, –5).

5 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

$\begin{array}{l} (i) L o c a t i o n o f P (1, - 1, 3) = (x, - y, z) = I Voctant \\ (i i) L o c a t i o n o f Q (- 1, 2, 4) = (- x, y, z) = I Ioctant \\ (i i i) L o c a t i o n o f R (- 2, - 4, - 7) = (- x, - y, - z) = V I Ioctant \\ (i v) L o c a t i o n o f S (- 4, 2, - 5) = (- x, y, - z) = V I octant \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of a circle that passes through (3, 6) and touches the axes is

(a) $x^{2} + y^{2} + 6 x + 6 y + 3 = 0$
(b) $x^{2} + y^{2} - 6 x - 6 y - 9 = 0$
(c) $x^{2} + y^{2} - 6 x - 6 y + 9 = 0$
(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:

$\begin{array}{l} L e t t h e r e q u i r e d c i r c l e t o u c h t h e a x e s a t (a, 0) a n d (0, a) \\ ∴ C e n t r e i s (a, a) a n d r = a \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - a)}^{2} + {(y - a)}^{2} = a^{2} \\ I f i t p a s s e s t h r o u g h a point P (3, 6) t h e n \\ {(3 - a)}^{2} + {(6 - a)}^{2} = a^{2} \\ \Rightarrow 9 + a^{2} - 6 a + 36 + a^{2} - 12 a = a^{2} \\ \Rightarrow a^{2} - 18 a + 45 = 0 \\ \Rightarrow a^{2} - 15 a - 3 a + 45 = 0 \\ \Rightarrow a (a - 15) - 3 (a - 15) = 0 \\ \Rightarrow (a - 3) (a - 15) = 0 \\ \Rightarrow a = 3 a n d a = 15 w h i c h i s n o t p o s s i b l e \\ ∴ a = 3 \\ S o, t h e r e q u i r e d e q u a t i o n o f t h e c i r c l e i s \\ {(x - 3)}^{2} + {(y - 3)}^{2} = 9 \\ \Rightarrow x^{2} + 9 - 6 x + y^{2} + 9 - 6 y = 9 \\ \Rightarrow x^{2} + y^{2} - 6 x - 6 y + 9 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The area of the circle centered at (1, 2) and passing through (4, 6) is

(a) $5 π$

(b) $10 π$

(c) $25 π$

(d) None of these

5 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:

$\begin{array}{l} G i v e n t h a t t h e c e n t r e o f t h e c i r c l e i s (1, 2) \\ R a d i u s o f t h e c i r c l e = \sqrt[]{{(4 - 1)}^{2} + {(6 - 2)}^{2}} \\ = \sqrt[]{9 + 16} = 5 \\ S o, t h e a r e a o f c i r c l e = π r^{2} = π {(5)}^{2} = 25 π \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the hyperbola with vertices at $(0, \pm 6)$ , eccentricity $\frac{5}{3}$ , and foci at $(0, \pm 10)$ is ______

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t e q u a t i o n o f t h e h y p e r b o l a i s - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ V e r t i c e s a r e (0, \pm b) ∴ b = 6 a n d e = \frac{5}{3} \\ W e k n o w t h a t e = \sqrt[]{1 + \frac{a^{2}}{b^{2}}} \\ \frac{5}{3} = \sqrt[]{1 + \frac{a^{2}}{36}} \Rightarrow \frac{25}{9} = 1 + \frac{a^{2}}{36} \\ \Rightarrow \frac{a^{2}}{36} = \frac{25}{9} - 1 = \frac{16}{9} \Rightarrow a^{2} = \frac{16}{9} \times 36 \\ \Rightarrow a^{2} = 64 \\ S o, t h e e q u a t i o n o f t h e h y p e r b o l a i s \\ - \frac{x^{2}}{64} + \frac{y^{2}}{36} = 1 \Rightarrow \frac{y^{2}}{36} - \frac{x^{2}}{64} = 1 \\ a n d f o c i = (0, \pm b e) = (0, \pm 6 \times \frac{5}{3}) = (0, \pm 10) \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{y^{2}}{36} - \frac{x^{2}}{64} = 1 a n d (0, \pm 10) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

The equation of the parabola having focus at (â€“1, â€“2) and the directrix $x - 2 y + 3 = 0$ is ______.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e a n y point o n t h e p a r a b o l a . \\ A c c o r d i n g t o t h e d e f i n i t i o n o f t h e p a r a b o l a \\ \sqrt[]{{(x_{1} + 1)}^{2} + {(y_{1} + 2)}^{2}} = | \frac{x_{1} - 2 y_{1} + 3}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2}}} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ x_{1}^{2} + 1 + 2 x_{1} + y_{1}^{2} + 4 + 4 y_{1} = \frac{x_{1}^{2} + 4 y_{1}^{2} + 9 - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1}}{5} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} + 2 x_{1} + 4 y_{1} + 5 = \frac{x_{1}^{2} + 4 y_{1}^{2} - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1} + 9}{5} \\ \Rightarrow 5 x_{1}^{2} + 5 y_{1}^{2} + 10 x_{1} + 20 y_{1} + 25 = x_{1}^{2} + 4 y_{1}^{2} - 4 x_{1} y_{1} - 12 y_{1} + 6 x_{1} + 9 \\ \Rightarrow 4 x_{1}^{2} + y_{1}^{2} + 4 x_{1} + 32 y_{1} + 4 x_{1} y_{1} + 16 = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 x^{2} + 4 x y + y^{2} + 4 x + 32 y + 16 = 0 . \end{array}$

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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 having foci (0, 1), (0, â€“1), and a minor axis of length 1 is ______.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Eleven

An ellipse is described by using an endless string that passes over two pins. If the axes are 6 cm and 4 cm, the length of the string and the distance between the pins are ______.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e e q u a t i o n o f a n e l l i p s e i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ H e r e, 2 a = 6 \Rightarrow a = 3 \\ a n d 2 b = 4 \Rightarrow b = 2 \\ W e k n o w t h a t c^{2} = a^{2} - b^{2} \\ = {(3)}^{2} - {(2)}^{2} = 9 - 4 = 5 \\ ∴ c = \sqrt[]{5}, w e h a v e e = \frac{c}{a} \Rightarrow e = \frac{\sqrt[]{5}}{3} \\ L e n g t h o f s t r i n g = 2 a + 2 a e = 2 a (1 + e) \\ = 6 (1 + \frac{\sqrt[]{5}}{3}) = \frac{6 (3 + \sqrt[]{5})}{3} = 6 + 2 \sqrt[]{5} \\ Distance b e t w e e n t h e p i n s = C C' = 2 a e = 2 \times 3 \times \frac{\sqrt[]{5}}{3} = 2 \sqrt[]{5} \\ H e n c e, t h e v a l u e o f f i l l e r a r e 6 + 2 \sqrt[]{5} c m a n d 2 \sqrt[]{5} c m . \end{array}$

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