The value value of `lambda` so that the line `y=2x+lambda` may touch the ellipse `3x^(2)+5y^(2)=15`

Options

A. `k=sqrt(23)`
B. `k=-sqrt(23)`
C. `k=+-sqrt(23)`
D. None of these

Answer

To find the value of \( \lambda \) such that the line \( y = 2x + \lambda \) touches the ellipse given by the equation \( 3x^2 + 5y^2 = 15 \), we will follow these steps:

### Step 1: Rewrite the ellipse equation in standard form
The given ellipse equation is:
\[ 3x^2 + 5y^2 = 15 \]
To convert this into standard form, we divide the entire equation by 15:
\[ \frac{3x^2}{15} + \frac{5y^2}{15} = 1 \]
This simplifies to:
\[ \frac{x^2}{5} + \frac{y^2}{3} = 1 \]
Here, we identify \( a^2 = 5 \) and \( b^2 = 3 \).

### Step 2: Identify the slope of the tangent line
The line \( y = 2x + \lambda \) has a slope \( m = 2 \).

### Step 3: Use the tangent line formula for ellipses
For an ellipse in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the equation of the tangent line at a point with slope \( m \) is given by:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Substituting \( a^2 = 5 \), \( b^2 = 3 \), and \( m = 2 \):
\[ y = 2x \pm \sqrt{5 \cdot 2^2 + 3} \]
Calculating the expression under the square root:
\[ = 5 \cdot 4 + 3 = 20 + 3 = 23 \]
Thus, the tangent line equation becomes:
\[ y = 2x \pm \sqrt{23} \]

### Step 4: Compare the tangent line equations
We have:
\[ y = 2x + \lambda \]
and from the tangent line formula:
\[ y = 2x \pm \sqrt{23} \]
To find \( \lambda \), we compare:
\[ \lambda = \pm \sqrt{23} \]

### Step 5: Conclusion
Thus, the values of \( \lambda \) for which the line touches the ellipse are:
\[ \lambda = \sqrt{23} \quad \text{or} \quad \lambda = -\sqrt{23} \]