The equation of the normal to the hyperbola ` (x^(2))/(25) -(y^(2))/(9) =1` at the point ` theta = (pi)/(4) ` is

The equation of the tangents to the hyperbola (x^(2))/( 9) -( y^(2))/( 4) = 1 at the point theta =(pi)/(3) is

The equation to the normal to the hyperbola (x^(2))/(16)-(y^(2))/(9)=1 "at "(-4,0) is

Find the equation of normal to the hyperbola (x^(2))/(16)-(y^(2))/(9)=1 at (-4,0)

Find the equation of the normal to the hyperbola x^(2)-y^(2)=9 at the point p(5,4).

The equation of the normal to the ellipse (x^(2))/(4)+(y^(2))/(2)=1 at the point whose eccentric angle is (pi)/(4) is

The equation of the normal to the hyperbola x^2/16-y^2/9=1 at point (8,3sqrt3) is :

The equation of the tangent to the hyperola x^(2)/9-y^(2)/4=1 at the point theta=pi/3 is

The equation of the tangent to the hyperola x^(2)/9-y^(2)/4=1 at the point theta=pi/3 is

The slop of the normal to the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 at the point ( a sec theta , b tan theta) is -