Let the eccentricity of the ellipse 2 x^2+a y^2-8 x-2 a y+(8-a)=0 be 1 3 . If the major axis of this ellipse is parallel to Y-axis, then the equation of the tangent to this ellipse with slope 1 is

Step-by-step solution

aligned & 2 x^2+a y^2-8 x-2 a y+8-a=0.....(i) & (2 x^2-8 x )+ (a y^2-2 a y )=a-8 & (x-2)^2 a + (y-1)^2 2 =1 aligned We have b^2=a^2 (i-e^2 ) and given that major axis is parallel to y -axis and e = 1 3 a=2 (1- 1 3 ) a= 4 3 Since slope of tangents d y d x =1 Now, 4 x+2 a y d y d x -8-2 a d y d x =0 Put d y d x =1, a= 4 3 , we get 3 x+2 y-8=0 y= 8-3 x 2 putting in (i), we get aligned & 15 x^2-60 x+52=0 & x=2 4 30 and y=1 6 30 aligned Equation of tangent is x-y-1 10 3 =0