Equation of a tangent to the ellipse `x^2/36+y^2/25=1`, passing through the point where a directrix of the ellipse meets the + ve x-axis is

Options

A. `5y+sqrt11x=36`
B. `6y+sqrt11x=25`
C. `6y+sqrt11x=36`
D. `6y-sqrt11x=36`

Answer

To find the equation of the tangent to the ellipse \( \frac{x^2}{36} + \frac{y^2}{25} = 1 \) that passes through the point where a directrix of the ellipse meets the positive x-axis, we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse can be compared to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where:
- \( a^2 = 36 \) implies \( a = 6 \)
- \( b^2 = 25 \) implies \( b = 5 \)

### Step 2: Calculate the eccentricity (e) of the ellipse
The eccentricity \( e \) is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{36}} = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6} \]

### Step 3: Find the coordinates of the point where the directrix meets the positive x-axis
The equation of the directrix for an ellipse is given by \( x = \frac{a}{e} \). Thus, we calculate:
\[ \frac{a}{e} = \frac{6}{\frac{\sqrt{11}}{6}} = \frac{6 \cdot 6}{\sqrt{11}} = \frac{36}{\sqrt{11}} \]
This point is \( \left(\frac{36}{\sqrt{11}}, 0\right) \).

### Step 4: Write the equation of the tangent line
The equation of the tangent to the ellipse at point \( (x_1, y_1) \) can be expressed as:
\[ yy_1 = \frac{b^2}{a^2} (x + x_1) \]
However, we need to express it in terms of the slope \( m \). The tangent line at a point on the ellipse can also be written as:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Since the tangent passes through \( \left(\frac{36}{\sqrt{11}}, 0\right) \), we substitute \( y = 0 \) and \( x = \frac{36}{\sqrt{11}} \):
\[ 0 = m \cdot \frac{36}{\sqrt{11}} \pm \sqrt{36m^2 + 25} \]

### Step 5: Solve for m
Rearranging gives:
\[ \sqrt{36m^2 + 25} = m \cdot \frac{36}{\sqrt{11}} \]
Squaring both sides:
\[ 36m^2 + 25 = \frac{1296m^2}{11} \]
Multiplying through by 11 to eliminate the fraction:
\[ 396m^2 + 275 = 1296m^2 \]
Rearranging:
\[ 900m^2 = 275 \implies m^2 = \frac{275}{900} = \frac{11}{36} \]
Thus, \( m = \pm \frac{\sqrt{11}}{6} \).

### Step 6: Write the final equation of the tangent
Using the point-slope form of the line equation:
\[ y - 0 = m\left(x - \frac{36}{\sqrt{11}}\right) \]
Substituting \( m = \frac{\sqrt{11}}{6} \) and simplifying:
\[ y = \frac{\sqrt{11}}{6} \left(x - \frac{36}{\sqrt{11}}\right) \]
This gives:
\[ y = \frac{\sqrt{11}}{6}x - 6 \]
For \( m = -\frac{\sqrt{11}}{6} \):
\[ y = -\frac{\sqrt{11}}{6} \left(x - \frac{36}{\sqrt{11}}\right) \]
This gives:
\[ y = -\frac{\sqrt{11}}{6}x + 6 \]

### Final Tangent Equation
Combining both tangents, we can express them in standard form:
1. \( \sqrt{11}x + 6y = 36 \) (for positive slope)
2. \( -\sqrt{11}x + 6y = 36 \) (for negative slope)

The equation of the tangent line to the ellipse passing through the point where the directrix meets the positive x-axis is:
\[ \sqrt{11}x + 6y = 36 \]