Question:

The portion of the tangent to the curve x\(^{\frac{2}{3}}\)+y\(^{\frac{2}{3}}\)=a\(^{\frac{2}{3}}\), a>0 at any point of it, intercepted between the axes

Updated On: Apr 26, 2025
  • varies at abscissa
  • varies as ordinate
  • is constant
  • varies as the product of abscissa and ordinate
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The Correct Option is C

Solution and Explanation

Step 1: Parametrize the curve. 

The equation of the curve is \( x^{2/3} + y^{2/3} = a^{2/3} \). We can parametrize it using \( x = a\cos^3\theta \) and \( y = a\sin^3\theta \).

Step 2: Find dy/dx.

First, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):

\[ \frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta \]

\[ \frac{dy}{d\theta} = 3a\sin^2\theta\cos\theta \]

Now, find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\frac{\sin\theta}{\cos\theta} = -\tan\theta \]

Step 3: Find the equation of the tangent line.

The equation of the tangent line at the point \( (a\cos^3\theta, a\sin^3\theta) \) is:

\[ y - a\sin^3\theta = -\tan\theta (x - a\cos^3\theta) \]

\[ y - a\sin^3\theta = -\frac{\sin\theta}{\cos\theta} (x - a\cos^3\theta) \]

\[ y\cos\theta - a\sin^3\theta\cos\theta = -x\sin\theta + a\cos^3\theta\sin\theta \]

\[ x\sin\theta + y\cos\theta = a\sin^3\theta\cos\theta + a\cos^3\theta\sin\theta \]

\[ x\sin\theta + y\cos\theta = a\sin\theta\cos\theta(\sin^2\theta + \cos^2\theta) \]

\[ x\sin\theta + y\cos\theta = a\sin\theta\cos\theta \]

Step 4: Find the x and y intercepts.

To find the x-intercept, set \( y = 0 \):

\[ x\sin\theta = a\sin\theta\cos\theta \]

\[ x = a\cos\theta \]

To find the y-intercept, set \( x = 0 \):

\[ y\cos\theta = a\sin\theta\cos\theta \]

\[ y = a\sin\theta \]

Step 5: Calculate the length of the intercept.

The length of the tangent intercepted between the axes is the distance between the points \( (a\cos\theta, 0) \) and \( (0, a\sin\theta) \), which is:

\[ L = \sqrt{(a\cos\theta - 0)^2 + (0 - a\sin\theta)^2} = \sqrt{a^2\cos^2\theta + a^2\sin^2\theta} = \sqrt{a^2(\cos^2\theta + \sin^2\theta)} = \sqrt{a^2} = a \]

Final Answer

Since \( L = a \), the length of the tangent intercepted between the axes is constant.

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Concepts Used:

Tangents and Normals

  • A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
  • Given a function y = f(x), the equation of the tangent for this curve at x = x0 
  • Slope of tangent (at x=x0) m=dy/dx||x=x0
  • A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got 

m×n = -1

  • The normal to a given curve y = f(x) at a point x = x0
  • The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m

Diagram Explaining Tangents and Normal: