Consider the particle travelling clockwise on the elliptical path `x^2/100+y^2/25=1` The particle leaves the orbit at the point `(-8, 3)` and travels in a straight line tangent to the ellpse. At what point will the particle cross the y-axis?

Options

A. `(0,-(25)/3)`
B. `(0,(7)/3)`
C. `(0,(25)/3)`
D. `(0,9)`

Answer

To solve the problem, we need to find the point where the particle crosses the y-axis after leaving the elliptical path defined by the equation \( \frac{x^2}{100} + \frac{y^2}{25} = 1 \) at the point \((-8, 3)\). We will follow these steps:

### Step 1: Write the equation of the ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{100} + \frac{y^2}{25} = 1 \]

### Step 2: Identify the point of tangency
The particle leaves the ellipse at the point \((-8, 3)\).

### Step 3: Use the point-slope form to find the equation of the tangent line
The equation of the tangent line to the ellipse at a point \((x_1, y_1)\) can be expressed as:
\[ \frac{x_1 x}{100} + \frac{y_1 y}{25} = 1 \]
Substituting \((-8, 3)\) into the equation:
\[ \frac{-8x}{100} + \frac{3y}{25} = 1 \]

### Step 4: Simplify the tangent equation
Multiplying through by 100 to eliminate the denominators:
\[ -8x + 12y = 100 \]
Rearranging gives:
\[ 12y - 8x = 100 \]
Dividing the entire equation by 4 to simplify:
\[ 3y - 2x = 25 \]

### Step 5: Find the intersection with the y-axis
To find where the particle crosses the y-axis, we set \(x = 0\) in the tangent equation:
\[ 3y - 2(0) = 25 \]
This simplifies to:
\[ 3y = 25 \]
Dividing both sides by 3 gives:
\[ y = \frac{25}{3} \]

### Step 6: Write the final point
The point where the particle crosses the y-axis is:
\[ (0, \frac{25}{3}) \]

### Final Answer
The particle will cross the y-axis at the point \((0, \frac{25}{3})\).

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