The x-intercept of the tangent at any arbitarary point of the curve `(a)/(x^(2))+(b)/(y^(2))=1` is proportioanl to

Options

A. square of the abscissa of the point of tangency
B. square root of the abscissa of the point of tangency
C. cube of the abscissa of the point of tangency
D. cube root of the abscissa of the point of tangency

Answer

To find the x-intercept of the tangent at any arbitrary point of the curve \(\frac{a}{x^2} + \frac{b}{y^2} = 1\), we will follow these steps:

### Step 1: Identify the arbitrary point
Let the arbitrary point on the curve be \(P(h, k)\).

### Step 2: Differentiate the curve
We start with the equation of the curve:
\[ \frac{a}{x^2} + \frac{b}{y^2} = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}\left(\frac{a}{x^2}\right) + \frac{d}{dx}\left(\frac{b}{y^2}\right) = 0 \]
Using the chain rule, we get:
\[ -\frac{2a}{x^3} - \frac{2b}{y^3} \frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the above equation gives:
\[ \frac{dy}{dx} = -\frac{a y^3}{b x^3} \]

### Step 4: Find the slope at point \(P(h, k)\)
Substituting \(x = h\) and \(y = k\) into the derivative:
\[ \frac{dy}{dx}\bigg|_{(h, k)} = -\frac{a k^3}{b h^3} \]

### Step 5: Write the equation of the tangent line
The equation of the tangent line at point \(P(h, k)\) can be expressed as:
\[ y - k = \left(-\frac{a k^3}{b h^3}\right)(x - h) \]

### Step 6: Find the x-intercept
To find the x-intercept, set \(y = 0\):
\[ 0 - k = \left(-\frac{a k^3}{b h^3}\right)(x - h) \]
Rearranging gives:
\[ x - h = \frac{b h^3}{a k^2} k \]
Thus,
\[ x = h + \frac{b h^3}{a k^2} \]

### Step 7: Simplify the expression
This can be rewritten as:
\[ x = \frac{b h^3}{a k^2} + h \]
This shows that the x-intercept is proportional to \(h^3\).

### Conclusion
The x-intercept of the tangent at any arbitrary point of the curve is proportional to \(h^3\).

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