The equation of the normal at `theta=pi/3` to the hyperbola `3x^2 – 4y^2 = 12 `is

Find the equation of the normal at theta=(pi)/(3) to the hyperbola 3x^(2)-4y^(2)=12 .

Find the equation of the normal at theta=(pi)/(3) to the hyperbola 3x^(2)-4y^(2)=12 .

Write down the equation of the normal at theta = (pi)/(3) to the hyperbola 3x^(2) - 4y^(2) = 12

The equation of the normal at theta=(pi)/(6) to the hyperbola 3 x^(2)-4 y^(2)=12 is

Find the equation of normal at point (4, 3) for the hyperbola 3x^(2) - 4y^(2) = 14 .

Find the equation of normal at point (4, 3) for the hyperbola 3x^(2) - 4y^(2) = 14 .

The equation of the normal to the hyperbola x^(2) -4y ^(2) =5 at (3,-1) is

The equation of the normal to the hyperbola x^(2) -4y ^(2) =5 at (3,-1) is

The equation of the tangent to the hyperbola 3x^(2) - 4y^(2) = 12 , which makes equal intercepts on the axes is

The equation of the tangent to the hyperbola 3x^(2) - 4y^(2) = 12 , which makes equal intercepts on the axes is