Find the condition that the curves; ax2 + by2 = 1 and a ‘x2 + b’ y2 = 1 may cut each other orthogonally (at right angles).


Correct option is

Condition for orthogonality implies that the tangents to the curves at the point of intersection are perpendicular. If (x0y0) is the point of intersection, and m1m2 are slopes of the tangents to the two curves a this point, them  

             m1m2 = –1.

Let us find the point of intersection. Solving the equations simultaneously, 

            ax2 + by2 – 1 = 0    

            a'x2 + b'y2 – 1 = 0   


⇒   the point of intersection (x0y0) is given by    


The slope of tangent to the curve ax2 + by2 = 1 is   


and the slope of tangent to the curve  


For orthogonality,   


Using the values of x0 and y0,    




If the line ax + by + c = 0 is a normal to the curve xy = 1 then


The point of intersection of the tangents drawn to the curve x2y = 1 – y at the points where it is meet by the curve xy = 1 – y is given by


The slope of the normal at the point with abscissa x = –2 of the graph of the function f (x) = | x2 – x | is


The tangent to the graph of the function y = f (x) at the point with abscissax = 1 form an angle of π/6 and at the point x = 2 an angle of π/3 and at the point x = 3 an angle of π/4. The value of 



The equations of the tangents to the curve y = x4 from the point (2, 0) not on the curve, are given by


The value of parameter a so that line (3 – ax + ay + (a2n – 1) = 0 is normal to the curve xy = 1, may lie in the interval


A cylindrical gas container is closed at the top and open at the bottom; if the iron plate forming the cylindrical sides. The ratio of the height to diameter of the diameter of the cylindrical using minimum material for the same capacity is


The critical points of the function f (x) where 


Find the abscissa of the point on the curve ay2 = x3, the normal at which cuts of equal intercept from the axes.


Find the locus of a point that divides a chord of slope 2 of the parabola y2= 4x internally in the ratio 1: 2.