Let `vec(r)` is a positive vector of a variable pont in cartesian OXY plane such that `vecr.(10hatj-8hati-vecr)=40` and `p_1=max{|vecr+2hati-3hatj|^2},p_2=min{|vecr+2hati-3hatj|^2}`. A tangent line is drawn to the curve `y=8/x^2` at the point A with abscissa 2. The drawn line cuts x-axis at a point B

Options

A. (a) 9
B. (b) `2sqrt2-1`
C. (c) `6sqrt6+3`
D. (d) `9-4sqrt2`

Explanation

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Understand the given vector equation
We are given the equation:
\[ \vec{r} \cdot (10\hat{j} - 8\hat{i} - \vec{r}) = 40 \]
Let \(\vec{r} = x\hat{i} + y\hat{j}\).

### Step 2: Expand the dot product
Substituting \(\vec{r}\) into the equation:
\[ (x\hat{i} + y\hat{j}) \cdot (10\hat{j} - 8\hat{i} - (x\hat{i} + y\hat{j})) = 40 \]
This simplifies to:
\[ (x\hat{i} + y\hat{j}) \cdot (10\hat{j} - 8\hat{i} - x\hat{i} - y\hat{j}) = 40 \]
\[ = (x\hat{i} + y\hat{j}) \cdot (-8 - x)\hat{i} + (10 - y)\hat{j} \]
Calculating the dot product:
\[ = x(-8 - x) + y(10 - y) = 40 \]
\[ -8x - x^2 + 10y - y^2 = 40 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 + 8x - 10y + 40 = 0 \]

### Step 4: Completing the square
Completing the square for \(x\) and \(y\):
\[ (x^2 + 8x + 16) + (y^2 - 10y + 25) = 1 \]
This simplifies to:
\[ (x + 4)^2 + (y - 5)^2 = 1 \]
This represents a circle with center at \((-4, 5)\) and radius \(1\).

### Step 5: Finding \(p_1\) and \(p_2\)
Next, we need to find \(p_1\) and \(p_2\):
\[ p_1 = \max \{ |\vec{r} + 2\hat{i} - 3\hat{j}|^2 \} \]
\[ p_2 = \min \{ |\vec{r} + 2\hat{i} - 3\hat{j}|^2 \} \]

Calculating:
\[ \vec{r} + 2\hat{i} - 3\hat{j} = (x + 2)\hat{i} + (y - 3)\hat{j} \]
The magnitude squared is:
\[ | \vec{r} + 2\hat{i} - 3\hat{j} |^2 = (x + 2)^2 + (y - 3)^2 \]

### Step 6: Finding maximum and minimum values
Since \((x + 4)^2 + (y - 5)^2 = 1\), we can express \(y\) in terms of \(x\):
\[ y = 5 + \sqrt{1 - (x + 4)^2} \quad \text{and} \quad y = 5 - \sqrt{1 - (x + 4)^2} \]
Substituting these into the expression for \(p_1\) and \(p_2\) will yield the maximum and minimum values.

### Step 7: Tangent to the curve
The curve given is \(y = \frac{8}{x^2}\). The derivative at \(x = 2\) gives the slope of the tangent line:
\[ \frac{dy}{dx} = -\frac{16}{x^3} \Rightarrow \text{At } x = 2, \frac{dy}{dx} = -2 \]
The equation of the tangent line at point \(A(2, 2)\) is:
\[ y - 2 = -2(x - 2) \Rightarrow y = -2x + 6 \]

### Step 8: Finding intersection with x-axis
Setting \(y = 0\):
\[ 0 = -2x + 6 \Rightarrow 2x = 6 \Rightarrow x = 3 \]
Thus, point \(B\) is \((3, 0)\).

### Conclusion
The coordinates of point \(B\) where the tangent line intersects the x-axis is \((3, 0)\).