if the tangent of the curve `y=(x)/(x^(2)-3)`,
`x in R,(xne+-sqrt(3))` at a point `(alpha,beta)ne(0,0)` on it is parallel to the line `2x+6y-11=0` then

if the tangent of the curve y=(x)/(x^(2)-3)x in R,(x!=+-sqrt(3)) at a point (alpha,beta)!=(0,0) on it is parallel to the line 2x+6y-11=0 then

If the tangent to the curve y= (x)/(x^(2) - 3), x in r ( xne pm sqrt3) , at a point (alpha, beta ) ne (0, 0) on it is parallel to the line 2x + 6y -11 =0 , then (A) |6alpha + 2beta|=19 (B) |6alpha + 2 beta|=9 (C) |2 alpha + 6 beta |= 19 (D) | 2 alpha + 6 beta| = 11

If the tangent to the curve y= (x)/(x^(2) - 3), x in r ( xne pm sqrt3) , at a point (alpha, beta ) ne (0, 0) on it is parallel to the line 2x + 6y -11 =0 , then (A) |6alpha + 2beta|=19 (B) |6alpha + 2 beta|=9 (C) |2 alpha + 6 beta |= 19 (D) | 2 alpha + 6 beta| = 11

If the tangent to the curve y= (x)/(x^(2) - 3), x in r ( xne pm sqrt3) , at a point (alpha, beta ) ne (0, 0) on it is parallel to the line 2x + 6y -11 =0 , then (a) |6α + 2β| = 9 (b) |2α + 6β| = 11 (c) |2α + 6β| = 19 (d) |6α + 2β| = 19

The equation of the tangent to the curve y={(x^2sin"1/x,xne0),(0,x=0):} at the origin is

The tangent to the curve x^2+y^2-2x-3=0 is parallel to X-axis at the points

If the tangent to the curve y=6x-x^2 is parallel to line 4x-2y-1=0, then the point of tangency on the curve is