=12x2−10x4 Therefore, the slope of the tangent at point (x,y) is 12x2−10x4. The equation of the tangent at (x,y) is given by Y−y=(12x2−10x4)(X−x)... (i) When, the tangent passes through the origin (0,0), then X=Y=0 Therefore, eq. (i) reduce to −y=(12x2−10x4)(−x) ⇒y=12x3−10x5 Also, we have y=4x3−2x5 ∴ 12x3−10x5 =4x3−2x5 ⇒ 8x5−8x3=0 ⇒x5−x3=0 ⇒ x3(x2−1)=0 ⇒x=0,±1 When, x=0, y=4(0)3−2(0)5=0 When, x=1, y=4(1)3−2(1)5=2 When, x=−1, y=4(−1)3−2(−1)5=−2 Hence, the require points are (0,0),(1,2) and (−1,−2).