The slope of the normal to the circle `x^(2)+y^(2)=a^(2)` at the point `(a cos theta, a sin theta)` is-

Find the equation of tangent to the circle x^(2)+y^(2)=a^(2) at the point (a cos theta, a sin theta) . Hence , show that the line y=x+a sqrt(2) touches the given circle,Find the coordinats of the point of contact.

The slope of tangent to the ellipse x^(2)+4y^(2)=4 at the point (2 cos theta, sin theta) "is" sqrt(2) , find the equation of the tangent.

The slope of tangent to the ellipse x^(2)+4y^(2)=4 at the point (2cos theta,sin theta) is sqrt(2) Find the equation of the tangent

If beta is one of the angles between the normals to the ellipse, x^2+3y^2=9 at the points (3 cos theta, sqrt(3) sin theta)" and "(-3 sin theta, sqrt(3) cos theta), theta in (0,(pi)/(2)) , then (2 cos beta)/(sin 2 theta) is equal to:

The slope of the normal to the curve : x=a(cos theta+theta sin theta), y =a(sin theta-theta sin theta) at any point 'theta' is :

The slope of the tangent to the curve r^(2)=a^(2)cos2 theta, where x=r cos theta and y=r sin theta, at the point theta=(pi)/(6) is

Find the slope of the normal to the curve x = 1 - a sin theta, y = b cos^2 theta at theta = pi/2

Find the slope of the normal to the curve x=a sin^(2) theta, y=b cos^(3)theta" at" theta=(pi)/(4) .

Find the slope of the normal to the curve x = a cos^(3) theta and y = a sin^(3) theta "at" theta = (pi)/(4)