Question 75566
The equation of the ellipse with center (h, k) is given by: 


{{{(x - h)^2/a^2}}} + {{{(y - k)^2/b^2}}} = 1 


Where the length of the major axis is greater than the minor axis. 


Here the centre is given by (8, 7)  and 


length of the major axis = 2a =  16 


lenght of the minor axis = 2b = 4 


This implies that a = 8 and b = 2  


Hence, substituting these in the above formula, we have: 


{{{(x - 8)^2/(8)^2}}} + {{{(y - 7)^2/(2)^2}}} = 1 


Simplifying these, we have: 


{{{(x - 8)^2/64}}} + {{{(y - 7)^2/4}}} = 1 


Taking LCM and simplifying this further, we have: 


{{{(x - 8)^2 + 16(y - 7)^2}}} = 64 


Simplifying this again, we have: 


{{{x^2 - 16x + 64 + 16(y^2 + 49 - 14y)}}} = 64  


{{{x^2 - 16x + 64 + 16y^2 + 784 - 224y}}} = 64 


{{{ x^2 - 16y^2 - 16x - 224y + 848}}} = 0 


This is the equation of the ellipse. 

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2. Write the equation of an ellipse with center (9, –3), horizontal major axis length 18, and minor axis length 10.


Solution:  


The equation of the ellipse with center (h, k) is given by: 


{{{(x - h)^2/a^2}}} + {{{(y - k)^2/b^2}}} = 1 


Where the length of the major axis is greater than the minor axis. 


Here the centre is given by (9, -3)  and 


length of the major axis = 2a =  18 


length of the minor axis = 2b = 10 


This implies that a = 9 and b = 5  


Hence, substituting these in the above formula, we have: 


{{{(x - 9)^2/(9)^2}}} + {{{(y - (-3))^2/(5)^2}}} = 1 


{{{(x - 9)^2/81}}} + {{{(y + 3)^2/25}}} = 1 


Taking LCM, we get: 


{{{25(x - 9)^2 + 81(y + 3)^2}}} = 2025 


{{{25(x^2 + 81 - 18y) + 81(y^2 + 9 - 6y)}}} = 2025 


{{{25x^2 + 2025 - 450x + 81y^2 + 729 + 486y}}} = 2025 


{{{25x^2 + 2025 - 450x + 81y^2 + 729 + 486y - 2025}}} = 0 


{{{25x^2 + 81y^2 - 450x + 486y + 729 = 0}}}