99 Percentile Qs Bank for JEE MainMathematicsEllipse

The equation of the ellipse having a vertex at (6,1) , a focus at (4,1) and the eccentricity 3 5 is

Options

A(x-1)^2 16 + (y-1)^2 25 =1
B(x-1)^2 25 + (y-1)^2 16 =1
C(x+1)^2 25 + (y+1)^2 16 =1
D(x+1)^2 16 + (y+1)^2 25 =1

Correct answer

B. (x-1)^2 25 + (y-1)^2 16 =1

Step-by-step solution

Let the equation of ellipse is, (x-h)^2 a^2 + (y-k)^2 b^2 =1 Now, array rlrl a-a e & =2 & a (1- 3 5 ) & =2 a=5 & So, & b & =4 array So, Now, vertex comparing the vertex, we are getting and aligned & 6-h=5 h=1 & 1-k=0 k=1 aligned So, equation of required ellipse is (x-1)^2 25 + (y-1)^2 16 =1 .

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