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New answer posted

a year ago

Class 11th 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).

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 11th 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.]

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th 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).

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

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

Class 11th 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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th 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).

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

Class 11th 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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th 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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 11th 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 ______

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th 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 ______.

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

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

Class 11th 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 ______.

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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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