If equation of directrix of an ellipse `x^2/a^2+y^2/b^2=1` is x=4, then normal to the ellipse at point `(1,beta),(beta gt 0)` passes through the point (where eccentricity of the ellipse is `1/2`)

Options

A. `(1,3/2)`
B. `(-1,3/2)`
C. `(-1,-3)`
D. `(3,-1)`

Answer

To solve the problem step by step, we will follow the logical flow as described in the video transcript.

### Step 1: Identify the given information
We are given the equation of an ellipse:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The directrix of the ellipse is given as \(x = 4\) and the eccentricity \(e = \frac{1}{2}\).

### Step 2: Use the directrix formula
The formula for the directrix of an ellipse is:
\[ x = \frac{a}{e} \]
Given that the directrix is \(x = 4\), we can equate:
\[ \frac{a}{e} = 4 \]
Substituting \(e = \frac{1}{2}\):
\[ \frac{a}{\frac{1}{2}} = 4 \implies a = 4 \cdot \frac{1}{2} = 2 \]

### Step 3: Calculate \(a^2\)
Now, we find \(a^2\):
\[ a^2 = 2^2 = 4 \]

### Step 4: Find \(b^2\) using eccentricity
We know the relationship between eccentricity, \(a^2\), and \(b^2\):
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting \(e = \frac{1}{2}\) and \(a^2 = 4\):
\[ \left(\frac{1}{2}\right)^2 = 1 - \frac{b^2}{4} \]
This simplifies to:
\[ \frac{1}{4} = 1 - \frac{b^2}{4} \]
Rearranging gives:
\[ \frac{b^2}{4} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus:
\[ b^2 = 3 \]

### Step 5: Write the equation of the ellipse
Now we can write the equation of the ellipse:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

### Step 6: Substitute the point \((1, \beta)\)
We substitute the point \((1, \beta)\) into the ellipse equation:
\[ \frac{1^2}{4} + \frac{\beta^2}{3} = 1 \]
This simplifies to:
\[ \frac{1}{4} + \frac{\beta^2}{3} = 1 \]
Rearranging gives:
\[ \frac{\beta^2}{3} = 1 - \frac{1}{4} = \frac{3}{4} \]
Multiplying by 3:
\[ \beta^2 = \frac{9}{4} \]
Thus:
\[ \beta = \frac{3}{2} \quad (\text{since } \beta > 0) \]

### Step 7: Find the slope of the tangent
Differentiate the ellipse equation implicitly:
\[ \frac{2x}{4} + \frac{2y \frac{dy}{dx}}{3} = 0 \]
This gives:
\[ \frac{dy}{dx} = -\frac{3}{4} \cdot \frac{x}{y} \]
At the point \((1, \frac{3}{2})\):
\[ \frac{dy}{dx} = -\frac{3}{4} \cdot \frac{1}{\frac{3}{2}} = -\frac{3}{4} \cdot \frac{2}{3} = -\frac{1}{2} \]

### Step 8: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = 2 \]

### Step 9: Write the equation of the normal
Using the point-slope form of the line:
\[ y - \frac{3}{2} = 2(x - 1) \]
Simplifying gives:
\[ y - \frac{3}{2} = 2x - 2 \implies 2x - y - \frac{1}{2} = 0 \implies 4x - 2y - 1 = 0 \]

### Step 10: Check which point satisfies the normal equation
We check the point \((1, \frac{3}{2})\):
\[ 4(1) - 2\left(\frac{3}{2}\right) - 1 = 4 - 3 - 1 = 0 \]
Thus, the point \((1, \frac{3}{2})\) lies on the normal.

### Final Answer:
The normal to the ellipse at the point \((1, \frac{3}{2})\) passes through the point.

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