Tangent is drawn at the point (-1,1) on the hyperbola `3x^2-4y^2+1=0`. The area bounded by the tangent and the coordinates axes is

Options

A. `(1)/(6)`
B. `(1)/(12)`
C. `(1)/(24)`
D. `(1)/(3)`

Answer

To find the area bounded by the tangent to the hyperbola \(3x^2 - 4y^2 + 1 = 0\) at the point \((-1, 1)\) and the coordinate axes, we will follow these steps:

### Step 1: Verify the point lies on the hyperbola
We need to check if the point \((-1, 1)\) satisfies the hyperbola equation.

\[ 3(-1)^2 - 4(1)^2 + 1 = 3(1) - 4(1) + 1 = 3 - 4 + 1 = 0 \]

Since the equation holds true, the point \((-1, 1)\) lies on the hyperbola.

### Step 2: Find the equation of the tangent line
The general equation of the tangent to the hyperbola \(3x^2 - 4y^2 = 1\) at the point \((x_0, y_0)\) is given by:

\[ 3xx_0 - 4yy_0 = 1 \]

Substituting \((-1, 1)\) into the equation:

\[ 3x(-1) - 4y(1) = 1 \]

This simplifies to:

\[ -3x - 4y = 1 \]

Rearranging gives:

\[ 3x + 4y = -1 \]

### Step 3: Find the x-intercept and y-intercept of the tangent line
To find the x-intercept, set \(y = 0\):

\[ 3x + 4(0) = -1 \implies 3x = -1 \implies x = -\frac{1}{3} \]

So, the x-intercept is \(\left(-\frac{1}{3}, 0\right)\).

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

\[ 3(0) + 4y = -1 \implies 4y = -1 \implies y = -\frac{1}{4} \]

So, the y-intercept is \(\left(0, -\frac{1}{4}\right)\).

### Step 4: Calculate the area of the triangle formed by the intercepts and the origin
The area \(A\) of the triangle formed by the intercepts and the origin can be calculated using the formula:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is the absolute value of the x-intercept and the height is the absolute value of the y-intercept:

\[ A = \frac{1}{2} \times \left(-\frac{1}{3}\right) \times \left(-\frac{1}{4}\right) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{24} \]

### Final Answer
The area bounded by the tangent and the coordinate axes is:

\[ \frac{1}{24} \text{ square units} \]

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