Question

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

Solution

Correct option is

x = 4a/9

The given curve is ay2 = x3                    …(i)   

Differentiate to get:   

             

The slope of normal    

Since the normal makes equal intercepts on the axes its inclinations to axis of x is either 45o or 135o.

So two normal are possible with slopes 1 and –1.    

   

One squaring 4a2y2 = 9x4. Using (i) we get:    

          4a x3 = 9x4        ⇒     x = 4a/9.

SIMILAR QUESTIONS

Q1

The distance between the origin and the normal to the curve

y = e2x + x2 at x = 0 is

Q2

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

Q3

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

Q4

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

Q5

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 

                          

Q6

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

Q7

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

Q8

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

Q9

The critical points of the function f (x) where 

Q10

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