The distance of the origin from the normal drawn at the point `(1,-1)` on the hyperbola `4x^2-3y^2=1` is

Options

A. `(3)/(5)`
B. `(4)/(5)`
C. `(1)/(5)`
D. `(7)/(5)`

Answer

To find the distance of the origin from the normal drawn at the point (1, -1) on the hyperbola given by the equation \(4x^2 - 3y^2 = 1\), we will follow these steps:

### Step 1: Convert the hyperbola equation to standard form
The given hyperbola is \(4x^2 - 3y^2 = 1\). We can rewrite this in standard form by dividing through by 1:

\[ \frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{3}} = 1 \]

This gives us \(a^2 = \frac{1}{4}\) and \(b^2 = \frac{1}{3}\).

### Step 2: Use the point form of the normal equation
The normal to the hyperbola at the point \((x_1, y_1)\) is given by the equation:

\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \]

Substituting \(x_1 = 1\), \(y_1 = -1\), \(a^2 = \frac{1}{4}\), and \(b^2 = \frac{1}{3}\):

\[ \frac{\frac{1}{4} x}{1} + \frac{\frac{1}{3} y}{-1} = \frac{1}{4} + \frac{1}{3} \]

### Step 3: Simplify the equation
This simplifies to:

\[ \frac{1}{4} x - \frac{1}{3} y = \frac{1}{4} + \frac{1}{3} \]

Finding a common denominator for the right side, we have:

\[ \frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \]

Thus, we have:

\[ \frac{1}{4} x - \frac{1}{3} y = \frac{7}{12} \]

### Step 4: Rearranging to standard line form
Multiplying through by 12 to eliminate fractions:

\[ 3x - 4y = 7 \]

This can be rearranged to:

\[ 3x - 4y - 7 = 0 \]

### Step 5: Calculate the perpendicular distance from the origin
The formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Here, \(A = 3\), \(B = -4\), \(C = -7\), and the point is the origin \((0, 0)\):

\[ d = \frac{|3(0) - 4(0) - 7|}{\sqrt{3^2 + (-4)^2}} = \frac{|-7|}{\sqrt{9 + 16}} = \frac{7}{\sqrt{25}} = \frac{7}{5} \]

### Final Answer
The distance of the origin from the normal drawn at the point (1, -1) on the hyperbola is:

\[ \frac{7}{5} \]