Find the tangents to the ellipse `x ^(2) + 9y ^(2) = 3, `which are (i) parallel (ii) perpendicular to the line `3x + 4y =9.`

Answer

To find the tangents to the ellipse \( x^2 + 9y^2 = 3 \) that are (i) parallel and (ii) perpendicular to the line \( 3x + 4y = 9 \), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse
The given ellipse can be rewritten in standard form:
\[ \frac{x^2}{3} + \frac{y^2}{\frac{1}{3}} = 1 \]
Here, we have \( a^2 = 3 \) and \( b^2 = \frac{1}{3} \).

### Step 2: Find the slope of the given line
The equation of the line is \( 3x + 4y = 9 \). To find the slope, we can rewrite it in slope-intercept form \( y = mx + c \):
\[ 4y = -3x + 9 \implies y = -\frac{3}{4}x + \frac{9}{4} \]
Thus, the slope \( m \) of the line is \( -\frac{3}{4} \).

### Step 3: Find the tangents parallel to the line
The slope of the tangents parallel to the line will also be \( -\frac{3}{4} \). The equation of the tangent to the ellipse can be written as:
\[ y = mx + c \]
Substituting \( m = -\frac{3}{4} \):
\[ y = -\frac{3}{4}x + c \]
Using the formula for the tangent to the ellipse:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \( m = -\frac{3}{4} \), \( a^2 = 3 \), and \( b^2 = \frac{1}{3} \):
\[ y = -\frac{3}{4}x \pm \sqrt{3 \left(-\frac{3}{4}\right)^2 + \frac{1}{3}} \]
Calculating:
\[ = -\frac{3}{4}x \pm \sqrt{3 \cdot \frac{9}{16} + \frac{1}{3}} = -\frac{3}{4}x \pm \sqrt{\frac{27}{16} + \frac{16}{48}} = -\frac{3}{4}x \pm \sqrt{\frac{81}{48}} \]
Thus, the tangents parallel to the line are:
\[ y = -\frac{3}{4}x \pm \frac{9}{4\sqrt{3}} \]

### Step 4: Find the slope of the tangents perpendicular to the line
For the tangents perpendicular to the line, the slope will be the negative reciprocal of \( -\frac{3}{4} \), which is \( \frac{4}{3} \).

### Step 5: Write the equation of the tangents perpendicular to the line
Using the slope \( \frac{4}{3} \):
\[ y = \frac{4}{3}x + c \]
Using the tangent formula:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \( m = \frac{4}{3} \):
\[ y = \frac{4}{3}x \pm \sqrt{3 \left(\frac{4}{3}\right)^2 + \frac{1}{3}} \]
Calculating:
\[ = \frac{4}{3}x \pm \sqrt{3 \cdot \frac{16}{9} + \frac{1}{3}} = \frac{4}{3}x \pm \sqrt{\frac{48}{9} + \frac{3}{9}} = \frac{4}{3}x \pm \sqrt{\frac{51}{9}} = \frac{4}{3}x \pm \frac{\sqrt{51}}{3} \]

### Final Results
1. The tangents parallel to the line \( 3x + 4y = 9 \) are:
\[ y = -\frac{3}{4}x \pm \frac{9}{4\sqrt{3}} \]
2. The tangents perpendicular to the line are:
\[ y = \frac{4}{3}x \pm \frac{\sqrt{51}}{3} \]