if the tangent at the point `(4 cos phi , (16)/(sqrt(11) )sin phi )` to the ellipse `16x^(2)+11y^(2) =256` Is also a tangent to the circle `x^(2) +y^(2)-2x=15,` then the value of `phi` is

Options

A. `+- pi//2`
B. `+-pi//4`
C. `+- pi//3`
D. `+- pi//6`

Answer

To solve the problem step by step, we will follow the mathematical reasoning laid out in the video transcript.

### Step 1: Write the equations of the ellipse and the circle

The given ellipse is:
\[ 16x^2 + 11y^2 = 256 \]

We can rewrite this in standard form:
\[ \frac{x^2}{16} + \frac{y^2}{\frac{256}{11}} = 1 \]
This means \( a^2 = 16 \) and \( b^2 = \frac{256}{11} \).

The given circle is:
\[ x^2 + y^2 - 2x = 15 \]

We can rewrite this as:
\[ (x - 1)^2 + y^2 = 16 \]
This indicates that the center of the circle is at \( (1, 0) \) and the radius is \( 4 \).

### Step 2: Find the equation of the tangent to the ellipse

The point on the ellipse is given as:
\[ (4 \cos \phi, \frac{16}{\sqrt{11}} \sin \phi) \]

Using the tangent formula for an ellipse:
\[ \frac{x \cdot x_1}{a^2} + \frac{y \cdot y_1}{b^2} = 1 \]
Substituting \( x_1 = 4 \cos \phi \) and \( y_1 = \frac{16}{\sqrt{11}} \sin \phi \):
\[ \frac{x \cdot (4 \cos \phi)}{16} + \frac{y \cdot \left(\frac{16}{\sqrt{11}} \sin \phi\right)}{\frac{256}{11}} = 1 \]

This simplifies to:
\[ \cos \phi \cdot x + \frac{\sqrt{11}}{4} \sin \phi \cdot y = 4 \]

### Step 3: Find the distance from the center of the circle to the tangent line

The center of the circle is \( (1, 0) \). The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For our tangent line:
\[ A = \cos \phi, B = \frac{\sqrt{11}}{4} \sin \phi, C = -4 \]

Substituting the center of the circle \( (1, 0) \):
\[ d = \frac{|\cos \phi \cdot 1 + 0 + (-4)|}{\sqrt{(\cos \phi)^2 + \left(\frac{\sqrt{11}}{4} \sin \phi\right)^2}} \]

Setting this equal to the radius of the circle, which is \( 4 \):
\[ \frac{|\cos \phi - 4|}{\sqrt{(\cos \phi)^2 + \frac{11}{16} \sin^2 \phi}} = 4 \]

### Step 4: Solve for \( \phi \)

Squaring both sides:
\[ (\cos \phi - 4)^2 = 16 \left((\cos \phi)^2 + \frac{11}{16} \sin^2 \phi\right) \]

Expanding and simplifying:
\[ \cos^2 \phi - 8 \cos \phi + 16 = 16 \cos^2 \phi + 11 \sin^2 \phi \]

Using \( \sin^2 \phi = 1 - \cos^2 \phi \):
\[ \cos^2 \phi - 8 \cos \phi + 16 = 16 \cos^2 \phi + 11(1 - \cos^2 \phi) \]
\[ \cos^2 \phi - 8 \cos \phi + 16 = 16 \cos^2 \phi + 11 - 11 \cos^2 \phi \]
\[ 0 = 4 \cos^2 \phi + 8 \cos \phi + 5 \]

This is a quadratic equation in \( \cos \phi \). Solving for \( \cos \phi \):
\[ 4 \cos^2 \phi + 8 \cos \phi + 5 = 0 \]

Using the quadratic formula:
\[ \cos \phi = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 - 80}}{8} \]
Since the discriminant is negative, we find that:
\[ \cos \phi = \frac{-8 \pm 2}{8} \]
This gives us:
\[ \cos \phi = -\frac{3}{4} \quad \text{or} \quad \cos \phi = -\frac{5}{4} \text{ (not possible)} \]

Thus, we have:
\[ \phi = \cos^{-1}\left(-\frac{3}{4}\right) \]

### Final Answer
The value of \( \phi \) is:
\[ \phi = \frac{5\pi}{3} \text{ or } \phi = \frac{2\pi}{3} \]