QUESTIONS BASED ON PARABOLA ELLIPSE AND HYPERBOLA

(1)  Find the equation of the parabola in each of the cases given below:

(i) focus (4, 0) and directrix x = −4.          Solution

(ii) passes through (2,-3) and symmetric about y -axis.        Solution

(iii) vertex (1,-2) and focus (4,-2).        Solution

(iv) end points of latus rectum(4,-8) and (4,8) .        Solution

(2)  Find the equation of the ellipse in each of the cases given below:

(i) foci (± 3 0), e  =  1/2            Solution

(ii) foci (0, ± 4) and end points of major axis are (0, ± 5).        Solution

(iii)  length of latus rectum 8, eccentricity = 3/5 and major axis on x -axis.        Solution

(iv) length of latus rectum 4 , distance between foci 4√2 and major axis as y - axis.        Solution

(3)  Find the equation of the hyperbola in each of the cases given below:

(i) foci (± 2, 0) , eccentricity = 3/2.        Solution

(ii) Centre (2, 1) , one of the foci (8, 1) and corresponding directrix x = 4 .       Solution

(iii) passing through (5,−2) and length of the transverse axis along x axis and of length 8 units.       Solution

(4)  Find the vertex, focus, equation of directrix and length of the latus rectum of the following:

(i) y²  = 16x             Solution

(ii) x² = 24y           Solution

(iii)  y² = −8x         Solution

(iv) x² - 2x + 8y + 17  =  0             Solution 

(v) y² - 4y - 8x + 12  =  0           Solution

(5)  Prove that the length of the latus rectum of the hyperbola (x²/a²) - (y²/b²)  =  1 is 2b²/a.       Solution

(6)  Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis         Solution

(7)  Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i)  (x²/25) + (y²/9) = 1            Solution

(ii)  (x²/3) + (y²/10) = 1        Solution

(iii)  (x²/25) - (y²/144) = 1        Solution

(iv)  (y²/16) - (x²/9) = 1        Solution

(8)  Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i)  [(x - 3)²/225] + [(y - 4)²/289] = 1          Solution

(ii)  [(x + 1)²/100] + [(y - 2)²/64] = 1        Solution

(iii)  [(x + 3)²/225] - [(y - 4)²/64] = 1        Solution

(iv)  [(y - 2)²/25] - [(x + 1)²/16] = 1        Solution

(v) 18x² + 12y² − 144x + 48y + 120 = 0        Solution

(vi) 9x² − y² − 36x − 6y + 18 = 0        Solution

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