The line (3 x+4 y-5=0 ) cuts the curve (2 x^2+3 y^2=5 ) at (A ) and (B ). If ' (O ) ' is the origin, then ( A O B= )

Step-by-step solution

Given equation of line is ( aligned 3 x+4 y-5 & =0 3 x+4 y 5 & =1 (i) aligned ) Equation of curve is (2 x^2+3 y^2=5 ) ...(ii) Homogenising Eq. (ii) using Eq. (i) ( aligned & 2 x^2+3 y^2=5(1)^2 & 2 x^2+3 y^2=5 ( 3 x+4 y 5 )^2 & 2 x^2+3 y^2=5 ( 9 x^2+16 y^2+24 x y 25 ) & 10 x^2+15 y^2=9 x^2+16 y^2+24 x y & x^2-y^2+24 x y=0 aligned ) Here, (coefficient of ( .x^2 )+ ( . ) coefficient of (y^2 ) ) ( aligned & =1+(-1)=0 A O B ! & =90^ (or) 2 aligned ) ( ) Hence, answer is (c).