Let a, b and `lambda` positive real numbers. Suppoose P is an end point of the latus rectum of the parabola `y^(2)=4 lambdax` and suppoose the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` passes thorugh the point. P. if the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is

Options

A. `(1)/(sqrt(2))`
B. `(1)/(2)`
C. `(1)/(3)`
D. `(2)/(5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the coordinates of point P
The point P is the endpoint of the latus rectum of the parabola given by the equation \( y^2 = 4\lambda x \). The coordinates of the endpoints of the latus rectum for this parabola are given by:
\[ P = (\lambda, 2\lambda) \]

### Step 2: Find the slope of the tangent to the parabola at point P
To find the slope of the tangent to the parabola at point P, we differentiate the equation of the parabola:
\[ y^2 = 4\lambda x \]
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 4\lambda \implies \frac{dy}{dx} = \frac{4\lambda}{2y} = \frac{2\lambda}{y} \]
Substituting \( y = 2\lambda \) (the y-coordinate of point P):
\[ \frac{dy}{dx} = \frac{2\lambda}{2\lambda} = 1 \]
Thus, the slope of the tangent to the parabola at point P is \( m_1 = 1 \).

### Step 3: Find the slope of the tangent to the ellipse at point P
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Differentiating this with respect to \( x \):
\[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{x}{y} \]
Substituting \( P = (\lambda, 2\lambda) \):
\[ \frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{\lambda}{2\lambda} = -\frac{b^2}{2a^2} \]
Let \( m_2 = -\frac{b^2}{2a^2} \).

### Step 4: Set the condition for perpendicular tangents
Since the tangents are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \implies 1 \cdot \left(-\frac{b^2}{2a^2}\right) = -1 \]
This simplifies to:
\[ -\frac{b^2}{2a^2} = -1 \implies \frac{b^2}{2a^2} = 1 \implies b^2 = 2a^2 \]

### Step 5: Find the eccentricity of the ellipse
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting \( b^2 = 2a^2 \):
\[ e = \sqrt{1 - \frac{a^2}{2a^2}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{1}{\sqrt{2}}} \]