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ELLIPSE (SOLVED PROBLEMS)

1. If (5, 12) and (24, 7) are the foci of an ellipse passing through the origin, then find the eccentricity of the ellipse.
A: If two foci are S(5, 12) and S'(24, 7) and the ellipse passes through origin O.


        

2. If the equation (5x - 1)2 + (5y - 2)2 = (λ2 - 2λ + 1) (3x + 4y - 1)2 represents an ellipse, then find values of 'λ'.
A: (5x - 1)2 + (5y - 2)2 = (λ- 1)2 (3x + 4y - 1)2


      

3. Find the equation of the ellipse whose axes are of length 6 and  and their equations are x - 3y + 3 = 0, 3x + y - 1 = 0 respectively
A: Equation of ellipse is


                                          

4. If PSQ is a focal chord of the ellipse 16x2 + 25y2 = 400 such that SP = 8, then find the length of SQ.
A:


          

5. An ellipse passes through the point (4, -1) and touches the line x + 4y - 10 = 0. Find its equation if its axes coincide with coordinate axes.

A: Let the equation of the ellipse be  
It passes through (4, -1) So a2 + 16b2 = a2b2  ------- (1) 
x + 4y - 10 = 0 touches the ellipse


                                                             

6. Find the angle between the pair of tangents from the point (1, 2) to the ellipse 3x2 + 2y2 = 5

A: The combined equation of the pair of tangents drawn from (1, 2) to the ellipse
    3x2 + 2y2 = 5
   (3x2 + 2y2 - 5) (3 + 8 - 5) = (3x + 4y - 5)2
   18x2 + 12y2 - 30 = 9x2 + 16y2 + 25 + 24xy - 30x - 40y = 0
    9x2 - 24xy - 16y2 + 30x + 40y - 55 = 0
   If the angle between these lines is θ, then tanθ =  
   where a = 9, h = -12, b = -4


  

 

7. If the ellipse     is inscribed in square of side length  'a' then a is equal to.
A: Since sides of the square are tangent and perpendicular to each other, so the vertices lie on director circle
 x2 + y2 = a2 - 7 + 13 - 5a
              = a2 - 5a + 6

given that  'a'  is side of the square
... a is the radius of the circum scribed circle
... a2 - 5a + 6 = a2


                            
But for an ellipse a2 - 7 > 0, 13 - 5a > 0

Hence no such a exists

Posted Date : 19-02-2021

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