If 4 x-3 y-5=0 is a normal to the ellipse 3 x^2+8 y^2=k , then the equation of the tangent drawn to this ellipse at the point (-2, m)(m 0) is

Step-by-step solution

Given the equation of ellipse 3 x^2+8 y^2=k x^2 k 3 + y^2 k 8 =1 Now, equation of normal to the ellipse at (x₁, y₁ ) is a^2 x₁ x- b^2 y₁ y=a^2-b^2 aligned & k 38 x₁ x- k 8 y₁ y= k 3 - k 8 & x 3 x₁ - y 8 y₁ = 5 24 8 x₁ x- 3 y₁ y=5 aligned Compare it with 4 x-3 y-5=0 We get (x₁, y₁ )=(2,1) Now, k=3 4+8=20 and 3 4+8 m^2=20 m=1, m 0 Since, 3 x^2+8 y^2=20 6 x+16 y y^ =0 (Differentiate both side) aligned & y^ = -6 x 16 y & at (-2,1), y^ = 12 16 = 3 4 aligned Now, equation of tangent at (-2,1) is aligned & y-1= 3 4 (x+2)