Find the equation of tangents to the ellipse `4x^(2)+5y^(2)=20` which are perpendicular to the line `3x+2y-5=0`

Answer

To find the equation of the tangents to the ellipse \(4x^2 + 5y^2 = 20\) that are perpendicular to the line \(3x + 2y - 5 = 0\), we can follow these steps:

### Step 1: Rewrite the equation of the ellipse
First, we rewrite the equation of the ellipse in standard form. The given equation is:
\[ 4x^2 + 5y^2 = 20 \]
Dividing the entire equation by 20, we get:
\[ \frac{x^2}{5} + \frac{y^2}{4} = 1 \]
This shows that \(a^2 = 5\) and \(b^2 = 4\).

### Step 2: Find the slope of the given line
Next, we convert the line equation \(3x + 2y - 5 = 0\) into slope-intercept form \(y = mx + c\):
\[ 2y = -3x + 5 \implies y = -\frac{3}{2}x + \frac{5}{2} \]
The slope \(m_1\) of this line is \(-\frac{3}{2}\).

### Step 3: Determine the slope of the tangent
Since we want the tangents to be perpendicular to this line, we can find the slope \(m_2\) of the tangent using the property that the product of the slopes of two perpendicular lines is \(-1\):
\[ m_1 \cdot m_2 = -1 \implies -\frac{3}{2} \cdot m_2 = -1 \implies m_2 = \frac{2}{3} \]

### Step 4: Use the tangent equation
The equation of the tangent to the ellipse at a point with slope \(m\) is given by:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
Substituting \(m = \frac{2}{3}\), \(a^2 = 5\), and \(b^2 = 4\):
\[ y = \frac{2}{3}x + \sqrt{5 \cdot \left(\frac{2}{3}\right)^2 + 4} \]
Calculating the term under the square root:
\[ = \sqrt{5 \cdot \frac{4}{9} + 4} = \sqrt{\frac{20}{9} + 4} = \sqrt{\frac{20}{9} + \frac{36}{9}} = \sqrt{\frac{56}{9}} = \frac{\sqrt{56}}{3} \]
Thus, the equation of the tangent becomes:
\[ y = \frac{2}{3}x + \frac{\sqrt{56}}{3} \]

### Step 5: Write the final equation of the tangent
Multiplying through by 3 to eliminate the fraction:
\[ 3y = 2x + \sqrt{56} \]
For the other tangent, we consider the negative square root:
\[ 3y = 2x - \sqrt{56} \]

### Final Result
The equations of the tangents to the ellipse that are perpendicular to the line are:
\[ 3y = 2x + \sqrt{56} \quad \text{and} \quad 3y = 2x - \sqrt{56} \]