The equation of the tangent to the ellipse `x^2+16y^2=16` making an angle of `60^(@)` with x-axis is

Options

A. `xsqrt(3)-y+7=0`
B. `xsqrt(3)-y-7=0`
C. `xsqrt(3)-y pm7=0`
D. None of these

Answer

To find the equation of the tangent to the ellipse \(x^2 + 16y^2 = 16\) that makes an angle of \(60^\circ\) with the x-axis, we can follow these steps:

### Step 1: Identify the slope of the tangent line
The slope \(m\) of the tangent line can be calculated using the tangent of the angle:
\[ m = \tan(60^\circ) = \sqrt{3} \]

### Step 2: Rewrite the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ x^2 + 16y^2 = 16 \]
Dividing the entire equation by 16, we get:
\[ \frac{x^2}{16} + \frac{y^2}{1} = 1 \]
This shows that \(a^2 = 16\) and \(b^2 = 1\), hence \(a = 4\) and \(b = 1\).

### Step 3: Use the formula for the equation of the tangent to the ellipse
The standard form of the equation of the tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at the point where the slope of the tangent is \(m\) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]

### Step 4: Substitute the values into the tangent equation
Substituting \(m = \sqrt{3}\), \(a^2 = 16\), and \(b^2 = 1\) into the tangent equation:
\[ y = \sqrt{3}x \pm \sqrt{16 \cdot 3 + 1} \]
Calculating the term under the square root:
\[ 16 \cdot 3 = 48 \quad \text{and} \quad 48 + 1 = 49 \]
Thus, we have:
\[ y = \sqrt{3}x \pm \sqrt{49} \]
\[ y = \sqrt{3}x \pm 7 \]

### Step 5: Write the final equation of the tangent
The final equations of the tangent lines are:
\[ y = \sqrt{3}x + 7 \quad \text{and} \quad y = \sqrt{3}x - 7 \]

### Step 6: Rearranging the equations
Rearranging these equations gives:
1. \(\sqrt{3}x - y + 7 = 0\)
2. \(\sqrt{3}x - y - 7 = 0\)

Thus, the equations of the tangents to the ellipse making an angle of \(60^\circ\) with the x-axis are:
\[ \sqrt{3}x - y \pm 7 = 0 \]