Question
The length of the portion of the tangent at any point on the curve x2/3 + y2/3 = a2/3 intercepted between the axis is
- a2
- a3
- a
- 2a2
The correct answer is: a
Related Questions to study
The equation of the tangent to the curve x2 + 2y = 8 which is the perpendicular to x – 2y + 1 = 0 is
For such questions, we should know the formula to find the tangent and slope of lines and curves.
The equation of the tangent to the curve x2 + 2y = 8 which is the perpendicular to x – 2y + 1 = 0 is
For such questions, we should know the formula to find the tangent and slope of lines and curves.
In an isosceles triangle the ends of the base are (2a, 0) and (0, a) and one side is parallel to x – axis. The third vertex is
In an isosceles triangle the ends of the base are (2a, 0) and (0, a) and one side is parallel to x – axis. The third vertex is
If x + 2y + 3 = 0, x + 2y – 7 = 0 and 2x – y – 4 = 0 form the sides of square, the equation of the fourth side is
For such questions, we should know properties of square. We should know the formula to find distance between two parallel lines.
If x + 2y + 3 = 0, x + 2y – 7 = 0 and 2x – y – 4 = 0 form the sides of square, the equation of the fourth side is
For such questions, we should know properties of square. We should know the formula to find distance between two parallel lines.
The number of real values of k for which the lines x – 2y + 3 = 0, kx + 3y + 1 = 0 and 4x – ky + 2 = 0 are concurrent is
For such questions, we should know properties of concurrent lines.
The number of real values of k for which the lines x – 2y + 3 = 0, kx + 3y + 1 = 0 and 4x – ky + 2 = 0 are concurrent is
For such questions, we should know properties of concurrent lines.
The distance of the line 2x – 3y = 4 from the point (1, 1) in the direction of the line x + y = 1 is
The distance of the line 2x – 3y = 4 from the point (1, 1) in the direction of the line x + y = 1 is
The mid points of the sides of a triangle are (5, 0), (5, 12) and (0, 12). The orthocentre of this triangle is
The mid points of the sides of a triangle are (5, 0), (5, 12) and (0, 12). The orthocentre of this triangle is
The image of the point A (1, 2) by the line mirror y = x is the point B and the image of B by line mirror y = 0 is the point
The image of the point A (1, 2) by the line mirror y = x is the point B and the image of B by line mirror y = 0 is the point
If 
For such questions, we should know different trigonometric formulas. We should simplify the function first before finding derivatives.
If 
For such questions, we should know different trigonometric formulas. We should simplify the function first before finding derivatives.
If 
For such questions, we should know the formulas of inverse functions.
If 
For such questions, we should know the formulas of inverse functions.
If 
For such questions, we should know different formulas.
If 
For such questions, we should know different formulas.
For such questions, we will use different formulas.
For such questions, we will use different formulas.
The length of the perpendicular from the incentre of the triangle formed by the axes and the line
to the hypotenuse is
The length of the perpendicular from the incentre of the triangle formed by the axes and the line
to the hypotenuse is
If
is the angle between two adjacent sides of a parallelogram and p, q are the distances between the parallel sides, then the area of the parallelogram
If
is the angle between two adjacent sides of a parallelogram and p, q are the distances between the parallel sides, then the area of the parallelogram
For such questions, we should know how to use u by v method.
For such questions, we should know how to use u by v method.
For such questions, we should know u.v method.
For such questions, we should know u.v method.