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Q1. Find the co-ordinates of the centre and the radius of the circle given by the equation x2 + y2 + 6x - 8y - 24 = 0.

Solution

Comparing the equation of the circle to the general equation of a circle x2 + y2 + 2gx + 2fy + c = 0.Centre of circle is (-g,-f)
Q2.

Solution

Q3. What is the eccentricity of the curve 4 x2 + y2 = 100?

Solution

E q u a t i o n space o f space t h e space e l l i p s e comma 4 x squared plus y squared equals 100 x squared over 25 plus y squared over 100 equals 1 a equals 5 comma space b equals 10 e squared equals 1 minus a squared over b squared rightwards double arrow e squared equals 1 minus 25 over 100 rightwards double arrow e squared equals 75 over 100 therefore e equals fraction numerator 5 square root of 3 over denominator 10 end fraction equals fraction numerator square root of 3 over denominator 2 end fraction
Q4. Find the equation of a circle whose coordinates of the end points of the diameter are (-3,2) and (2,-4).

Solution

The equation of the circle when the end points of the diameter are given is (x - ) (x - ) + (y - ) (y - ) = 0.Hence the equation is (x + 3) (x - 2) + (y - 2) (y + 4) = 0or x2 + x - 6 + y2 + 2y - 8 = 0or x2 + y2 - x + 2y - 14 = 0
Q5. A ladder 12 m long leaning against the wall begins to slide down. Its one end always remains on the wall and the other on the floor. Find the equation of the locus of a point P which is 3 m from the end in contact with the floor. Identify the conic section represented by the equation.

Solution

A c c o r d i n g space t o space t h e space q u e s t i o n A B equals 12 m P B equals 3 m A P equals 9 m F r o m space t h e space f i g u r e space w e space c a n space s a y space t h a t space p o i n t space P space d i v i d e s space A B space i n space t h e space r a t i o space 9 colon 3 equals 3 colon 1 C o o r d i n a t e s space o f space P space i s space g i v e n space b y x equals open parentheses fraction numerator 3. b plus 1.0 over denominator 4 end fraction close parentheses equals fraction numerator 3 b over denominator 4 end fraction rightwards double arrow b equals fraction numerator 4 x over denominator 3 end fraction y equals open parentheses fraction numerator 3.0 plus 1. a over denominator 4 end fraction close parentheses equals a over 4 rightwards double arrow a equals 4 y F r o m space t h e space t r i a n g l e space w e space c a n space w r i t e space a squared plus b squared equals 12 squared rightwards double arrow 16 y squared plus fraction numerator 16 x squared over denominator 9 end fraction equals 144 rightwards double arrow x squared over 81 plus y squared over 9 equals 1 space w h i c h space i s space t h e space e q u a t i o n space o f space e l l i p s e S o space t h e space c o n i c space i s space a n space e l l i p s e
Q6. The space number space of space common space tangents space to space the space circles space straight x squared plus straight y squared minus 4 straight x minus 6 straight y minus 12 equals 0 and space straight x squared plus straight y squared plus 6 straight x plus 18 straight y plus 26 equals 0 space is colon
  • 1) 3
  • 2) 2
  • 3) 4
  • 4) 1

Solution

Q7.

Solution

Q8. Find the equation of the circle with centre (-2,3) and radius 4.

Solution

Equation of a circle when the centre and radius are given is (x - h)2 + (y - k)2 = r2.Hence h = -2, k = 3 and r = 4. (x + 2)2 + (y - 3)2 = 42    (x + 2)2 + (y - 3)2 = 16.
Q9. Find the equation of the parabola with focus (0,4) and directix y = -4.

Solution

The focus lies on the y axis, so it is an upward or a downwoard parabola directix is y = -4. Hence it is an upward parabola x2 = 4ay.Where a = 4 Equation of the parabola is x2 = 4 4 y = 16y. x2 = 16y.
Q10. Find the focus of a parabola whose axis lies along the X- axis and the parabola passes through the point (5, 10)

Solution

T h e space p a r a b o l a space i s space g i v e n space b y space y squared equals 4 a x S i n c e space t h e space p o i n t space left parenthesis 5 comma space 10 right parenthesis space l i e s space o n space i t comma space t h e r e f o r e 10 squared equals 4 a open parentheses 5 close parentheses rightwards double arrow a equals 100 over 20 equals 5
Q11. Show that the points (5,5), (6,4) (-2,4) and (7,1) are concyclic. Find the equation, centre and radius of the circle.

Solution

Let the equation of the circle passing through the points (5,5), (6,4) and (-2,4) be x2 + y2 + 2gx + 2fy + c = 0 ...(i)The points (5,5), (6,4) and (-2,4) lie on the circle, so they must satisfy the equation (i)25 + 25 + 10g + 10f + c = 0 ...(ii)36 + 16 + 12g + 8f + c = 0 ....(iii)4 + 16 - 4g + 8f + c = 0 ...(iv)subtracting equations (iii) from (ii) and (iv) and (iii) we get   Error converting from MathML to accessible text.
Q12.

Solution

L e t space t h e space e q u a t i o n space o f space t h e space e l l i p s e space b e space x squared over a squared plus y squared over b squared equals 1 space open parentheses a greater than b close parentheses H e r e space V e r t i c e s space a r e space left parenthesis plus-or-minus 13 comma 0 right parenthesis space a n d space f o c i left parenthesis plus-or-minus 5 comma 0 right parenthesis a equals 13 a e equals 5 rightwards double arrow e equals 5 over 13 a squared e squared equals a squared minus b squared rightwards double arrow 25 equals 169 minus b squared rightwards double arrow b squared equals 144 rightwards double arrow b equals 12 therefore x squared over 13 squared plus y squared over 12 squared equals 1 rightwards double arrow x squared over 169 plus y squared over 144 equals 1
Q13. Find the equation of the ellipse with centre at (0,0) and passes through the point (4,3) and (6,2).

Solution

L e t space t h e space e q u a t i o n space o f space t h e space e l l i p s e space b e space x squared over a squared plus y squared over b squared equals 1 T h e space e l l i p s e space p a s s e s space t h r o u g h space t h e space p o i n t s space left parenthesis 4 comma 3 right parenthesis space a n d space left parenthesis 6 comma 2 right parenthesis space S o space w e space h a v e 16 over a squared plus 9 over b squared equals 1 space rightwards double arrow 16 b squared plus 9 a squared equals a squared b squared rightwards arrow circle enclose 1 a n d space 36 over a squared plus 4 over b squared equals 1 rightwards double arrow 36 b squared plus 4 a squared equals a squared b squared rightwards arrow circle enclose 2 S o l v i n g space circle enclose 1 space a n d space circle enclose 2 space w e space g e t a squared equals 52 space a n d space b squared equals 13 space S o space t h e space e q u a t i o n space o f space t h e space e l l i p s e space i s space x squared over 52 plus y squared over 13 equals 1
Q14. An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution

Q15. Find the equation of the circle which passes through the points (3,7), (5,5) and has its centre on the line x - 4y = 1.

Solution

Let the equation of the required circle be x2 + y2 + 2gx + 2fy + C =0..(i)It passes through (3,7) and (5,5) 9 + 49 + 6g + 14f + C = 0 ...(ii)25 + 25 + 10g + 10f + C = 0 ...(iii) The centre (-g, -f) lies on x - 4y = 1 -g + 4f = 1 ...(iv)Subtracting (ii) from (iii), we get 4g - 4f - 8 = 0. ...(v)or g - f = 2 ...(vi)Solving equation (iv) and (vi)3f = 3 ,  g - 1 = 2f = 1 ,  g = 3.Substituting the values of g and f in (ii) We get 9 + 49 + 18 + 14 + C = 0. C = -90Hence the equation of the circle x2 + y2 + 6x + 2y - 90 = 0.
Q16.

Solution

S i n c e space t h e space d e n o m i n a t o r space o f space x squared space i s space g r e a t e r space t h e n space y squared comma space t h e space m a j o r space a x i s space i s space a l o n g space X minus a x i s. C o m p a r i n g space t h e space t h e space s tan d a r d space e q u a t i o n space o f space e l l i p s e space x squared over a squared plus y squared over b squared equals 1 space w e space g e t a equals 5 space a n d space b equals 3 b squared equals a squared left parenthesis 1 minus e squared right parenthesis rightwards double arrow e squared equals 1 minus 9 over 25 equals 16 over 25 rightwards double arrow e equals 4 over 5 t h e space c o o r d i n a t e s space o f space f o c i i space a r e space left parenthesis plus-or-minus a e comma 0 right parenthesis equals left parenthesis plus-or-minus 4 comma 0 right parenthesis
Q17. Find the equation of the parabola with vertex (0,0), passing through the point (4,5) and symmetric about the x - axis.

Solution

Since the parabola is symmetric about the x axis, it is of the form y2 = 4ax.Hence it passes through the point (4,5), the co-ordinates should satisfy the equation of the parabolaHence replace x = 4, and y = 5 in the equation y2 = 4ax, we get 25 = 4 a 4 a = 25/16.Hence the equation of the parabola is y2 = 25x/16 or 16y2 = 25x


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