Show that curves `x^(3)-3xy^(2)+2=0 and y^(3)-3x^(2)y+2=0` intersect at right angles.

The ellipse 4x^2+9y^2=36 and the hyperbola a^2x^2-y^2=4 intersect at right angles. Then the equation of the circle through the points of intersection of two conics is (a) x^2+y^2=5 (b) sqrt(5)(x^2+y^2)-3x-4y=0 (c) sqrt(5)(x^2+y^2)+3x+4y=0 (d) x^2+y^2=25

The circles x^2+y^2+x+y=0 and x^2+y^2+x-y=0 intersect at an angle of

The circle x^2+y^2+x+y=0 and x^2+y^2+x-y=0 intersect at an angle of :

The line tangent to the curves y^3-x^2y+5y-2x=0 and x^2-x^3y^2+5x+2y=0 at the origin intersect at an angle theta equal to (a) pi/6 (b) pi/4 (c) pi/3 (d) pi/2

The normal to the curve, x^(2)+2xy-3y^(2)=0 at (1, 1)-

If the tangent to the curve x^(3)+y^(3)=a^(3) at the point (x_(1),y_(1)) intersects the curve again at the point (x_(2),y_(2)) , then show that, (x_(2))/(x_(1))+(y_(2))/(y_(1))+1=0 .

(x^(2)-2xy)dy+(x^(2)-3xy+2y^(2))dx=0

If the curves ax^2+4xy+2y^2+x+y+5=0 and ax^2+6xy+5y^2+2x+3y+8=0 intersect at four concyclic points then the value of a is

Prove that the tangent to the curce y=x^(2)-5x+6 at the points (2,0) and (3,0) are at right angles.

Show that the tangents to the curve y=x^2-5x+6 at the point (2,0) and (3,0) are at right angle.