Question

Find the equation of ellipse whose one vertex is (3 1)

Answer 1

Step-by-step explanation:

Given  

• Find the equation of ellipse whose one vertex is (3 1),the nearer focus is (1,1) and eccentricity is 2/3.
• We know the equation of the ellipse as x^2 / a^2 + y^2 / b^2 = 1
• In a graph vertex will be (-a,0) and (a,0) and focus will be s1 and s2.  
• So coordinates will be (ae,0), so s1v will be a – ae
• Eccentricity is 2/3  
• So e = 2/3 = √1 – b^2 / a^2
•               = 1 – 4/9
•              = 5/9
• So distance of SV = a – ae (ae = 1 since centre c = 2)
•                             = a(1 – e) = 2
•                             = a(1 – 2/3) = 2
•                             = a x 1/3 = 2
•                        So a = 6
• Now b^2 / a^2 = 5/9
• So b^2 = 5/9 x 36
• Now b^2 = 20 and a^2 = 36
• Now 1 – h = ae
•                    = 6 x 2/3
•                    = 4
• So 1 – h = 4
• Or h = - 3
• We have centre of coordinates of ellipse will be (-3,1)
• Now x = 1 and y = - 3 are the major and minor axis
• Therefore equation of ellipse will be (x – 1)^2 / 36 + (y + 3)^2 / 20 = 1

Reference link can be

https://brainly.in/question/7255356