If the tangent drawn at a point `(t^(2),2t)` on the parabola `y^(2)=4x` is same as normal drawn at `(sqrt(5)cosalpha, 2sinalpha)` on the ellipse `(x^(2))/(5)+(y^(2))/(4)=1`, then which of following is not true ?

Options

A. `t=pm(1)/(sqrt(5))`
B. `alpha=-tan^(-1)2`
C. `alpha=tan^(-1)2`
D. none

Answer

To solve the problem, we need to analyze the conditions given for the parabola and the ellipse.

### Step 1: Find the equation of the tangent to the parabola
The equation of the parabola is given by \( y^2 = 4x \). The point on the parabola is \( (t^2, 2t) \).

The equation of the tangent to the parabola at the point \( (t^2, 2t) \) is given by:
\[ yy_1 = 2(x + x_1) \]
Substituting \( (x_1, y_1) = (t^2, 2t) \):
\[ y(2t) = 2(x + t^2) \]
This simplifies to:
\[ 2ty = 2x + 2t^2 \implies ty = x + t^2 \implies x - ty + t^2 = 0 \]

### Step 2: Find the equation of the normal to the ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{5} + \frac{y^2}{4} = 1 \]
The point on the ellipse is \( \left(\sqrt{5} \cos \alpha, 2 \sin \alpha\right) \).

The equation of the normal to the ellipse at the point \( (x_1, y_1) \) is given by:
\[ \frac{a^2}{x_1^2}(x - x_1) + \frac{b^2}{y_1^2}(y - y_1) = 0 \]
Where \( a^2 = 5 \) and \( b^2 = 4 \). Substituting \( (x_1, y_1) = \left(\sqrt{5} \cos \alpha, 2 \sin \alpha\right) \):
\[ \frac{5}{5 \cos^2 \alpha}(x - \sqrt{5} \cos \alpha) + \frac{4}{4 \sin^2 \alpha}(y - 2 \sin \alpha) = 0 \]
This simplifies to:
\[ \frac{1}{\cos^2 \alpha}(x - \sqrt{5} \cos \alpha) + \frac{1}{\sin^2 \alpha}(y - 2 \sin \alpha) = 0 \]
Multiplying through by \( \cos^2 \alpha \sin^2 \alpha \):
\[ \sin^2 \alpha (x - \sqrt{5} \cos \alpha) + \cos^2 \alpha (y - 2 \sin \alpha) = 0 \]
Expanding this gives:
\[ \sin^2 \alpha x - \sqrt{5} \sin^2 \alpha \cos \alpha + \cos^2 \alpha y - 2 \cos^2 \alpha \sin \alpha = 0 \]

### Step 3: Set the equations equal
Now we have the equations for the tangent and the normal:
1. Tangent: \( x - ty + t^2 = 0 \)
2. Normal: \( \sin^2 \alpha x + \cos^2 \alpha y = \sqrt{5} \sin^2 \alpha \cos \alpha + 2 \cos^2 \alpha \sin \alpha \)

For these to be the same, their coefficients must be proportional.

### Step 4: Find the relationships
From the tangent equation, we can express \( t \) in terms of \( \alpha \):
\[ \frac{1}{t} = \frac{\sin^2 \alpha}{1} \quad \text{and} \quad \frac{1}{t^2} = \frac{\cos^2 \alpha}{1} \]
This leads to relationships between \( t \), \( \sin \alpha \), and \( \cos \alpha \).

### Step 5: Analyze the conditions
From the relationships derived, we can analyze which conditions are not true based on the values of \( t \) and \( \alpha \).

### Conclusion
After analyzing the derived equations and their relationships, we can conclude which statements about \( t \) and \( \alpha \) are not true based on the conditions set by the problem.