{"id":4621,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-9\/"},"modified":"2016-05-17T11:35:10","modified_gmt":"2016-05-17T06:05:10","slug":"ncert-solutions-class-11-maths-miscellaneous-9","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-9\/","title":{"rendered":"NCERT Solutions class-11 Maths Miscellaneous"},"content":{"rendered":"

Miscellaneous Exercise<\/strong><\/p>\n

1. In a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.<\/strong><\/p>\n

Ans. <\/strong>A parabolic reflector with diameter PR = 20 cm and OQ = 5 cm<\/p>\n

Vertex of parabola is (0, 0)<\/p>\n

Let focus of the parabola be <\/p>\n

<\/p>\n

Now, PR = 20 cm<\/p>\n

PQ = 10 cm<\/p>\n

Coordinate of the point P are (5, 10)<\/p>\n

Since the point lies on the parabola <\/p>\n

<\/p>\n

<\/p>\n

\n

2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?<\/strong><\/p>\n

Ans. <\/strong>Let AB be the parabolic arch having O as the vertex and OY as the axis.<\/p>\n

<\/p>\n

The parabola is of the form <\/p>\n

Now, CD = 5 m<\/p>\n

OD = 2.5 m<\/p>\n

And BD = 10 m<\/p>\n

Therefore, coordinates of point B are (2.5, 10).<\/p>\n

Since the point B lies on the parabola <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Equation of the parabola is <\/p>\n

<\/p>\n

Let PQ = <\/p>\n

NQ = <\/p>\n

Coordinate of the point Q are <\/p>\n

Since the point Q lies on the parabola <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, width of arch is m = 2.24 m approx.<\/p>\n

\n

3. The cable of a uniformly loaded suspension bridge hange in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.<\/strong><\/p>\n

Ans. <\/strong>Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the roadway.<\/p>\n

Now AL = BM = 30 m, OC = 6 m and LM = 100 m<\/p>\n

<\/p>\n

LC = CM = LM = 50 m<\/p>\n

Let O be the vertex and axis of the parabola be axis.<\/p>\n

Therefore, the equation of the parabola in standard form is <\/p>\n

Coordinates of the point B are (50, 24)<\/p>\n

Since point B lies on the parabola <\/p>\n

<\/p>\n

<\/p>\n

Therefore, equation of parabola is <\/p>\n

<\/p>\n

Let length of the supporting wire PQ at a distance of 18 m be <\/p>\n

OR = 18 m and PR = PQ \u2013 QR = PQ \u2013 OC = <\/p>\n

Coordinates of point P are <\/p>\n

Now, since the point P lies on parabola <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

9.11 m approx.<\/p>\n

\n

4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.<\/strong><\/p>\n

Ans. <\/strong>Given: Width of elliptical arch (AB)<\/p>\n

= = 8 = 4 m<\/p>\n

<\/p>\n

Height of the centre (OB) = = 2 m<\/p>\n

The axis of ellipse is axis.<\/p>\n

Therefore, the equation of ellipse in standard form is .<\/p>\n

<\/p>\n

<\/p>\n

Now, AP = 1.5 m<\/p>\n

OP = OA \u2013 AP = 4 \u2013 1.5 = 2.5 m<\/p>\n

Let PQ = , then coordinates of Q are <\/p>\n

Since the point Q lies on the ellipse <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

= 1.56 m approx.<\/p>\n

\n

5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the <\/strong>axis.<\/strong><\/p>\n

Ans. <\/strong>Let AB be a rod of length 12 cm and P be any point on the rod such that PA = 3 cm and PB = 9 cm.<\/p>\n

<\/p>\n

Let AR = and BQ = <\/p>\n

Then <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

And <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Now, OA = OR + AR = <\/p>\n

And OB = OQ + BQ = <\/p>\n

In right angled triangle AOB, <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

which is required locus of point P and which represents an ellipse.<\/p>\n

\n

6. Find the area of the triangle formed by the lines the vertex of the parabola <\/strong> to the ends of its latus rectum.<\/strong><\/p>\n

Ans. <\/strong>Given: Equation of parabola which is in the form of <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Focus of the parabola is (0, 3).<\/p>\n

Let AB be the latus rectum of the parabola, then <\/p>\n

<\/p>\n

<\/p>\n

The coordinates of A are <\/p>\n

and coordinates of B are (6, 3).<\/p>\n

Area of <\/p>\n

= <\/p>\n

= <\/p>\n

= = 18 sq. units<\/p>\n

\n

7. A man running a race leave no space course notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.<\/strong><\/p>\n

Ans. <\/strong>Let be two points where the flag parts are fixed on the ground. The origin O is the mid-point of.<\/p>\n

<\/p>\n

= = 4 m<\/p>\n

Coordinates of F1<\/sub> are and F2<\/sub> are (4, 0).<\/p>\n

Let P be any point on the track.<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Squaring both sides, we have,<\/p>\n

<\/p>\n

<\/p>\n

Squaring both sides again, we have,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

which is the required equation of locus of point P.<\/p>\n

\n

8. An equilateral triangle is inscribed n the parabola <\/strong> where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.<\/strong><\/p>\n

Ans. <\/strong>Given: Equation of the parabola .<\/p>\n

<\/p>\n

Let be the side of an equilateral whose one vertex<\/p>\n

is the vertex of parabola and let OC = <\/p>\n

Now, AB = <\/p>\n

AC = BC = = <\/p>\n

Coordinates of point A are x\u2019 x<\/p>\n

Since, point A lies on the parabola <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

In right angled triangle <\/p>\n

OA2<\/sup> = OC2<\/sup> + AC2<\/sup><\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the side of triangle is <\/p>\n","protected":false},"excerpt":{"rendered":"

Miscellaneous Exercise 1. In a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Ans. A parabolic reflector with diameter PR = 20 cm and OQ = 5 cm Vertex of parabola is (0, 0) Let focus of the parabola be Now, PR = 20 cm PQ = 10 cm … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47,281],"tags":[],"class_list":["post-4621","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Miscellaneous | myCBSEguide<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-9\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions class-11 Maths Miscellaneous | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Miscellaneous Exercise 1. In a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Ans. A parabolic reflector with diameter PR = 20 cm and OQ = 5 cm Vertex of parabola is (0, 0) Let focus of the parabola be Now, PR = 20 cm PQ = 10 cm ... 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