How many tangents to the circle `x^2 + y^2 = 3` are normal tothe ellipse `x^2/9+y^2/4=1?`

Options

A. 3
B. 2
C. 1
D. 0

Explanation

To solve the problem of finding how many tangents to the circle \(x^2 + y^2 = 3\) are normal to the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\), we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse is \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). Here, \(a^2 = 9\) and \(b^2 = 4\), which gives us \(a = 3\) and \(b = 2\).

### Step 2: Write the equation of the normal to the ellipse
The equation of the normal to the ellipse at a point \((a \cos \theta, b \sin \theta)\) is given by:
\[ x \sec \theta - y \cos \theta = a^2 - b^2 \]
Substituting \(a^2 = 9\) and \(b^2 = 4\), we have:
\[ x \sec \theta - y \cos \theta = 9 - 4 = 5 \]
Thus, the equation of the normal becomes:
\[ 3x \sec \theta - 2y \cos \theta = 5 \]

### Step 3: Find the perpendicular distance from the center of the circle to the normal line
The center of the circle \(x^2 + y^2 = 3\) is at the origin \((0, 0)\). The perpendicular distance \(d\) from the center to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]
Rearranging the normal equation:
\[ 3x \sec \theta - 2y \cos \theta - 5 = 0 \]
Here, \(A = 3 \sec \theta\), \(B = -2 \cos \theta\), and \(C = -5\). Thus, the distance is:
\[ d = \frac{|-5|}{\sqrt{(3 \sec \theta)^2 + (-2 \cos \theta)^2}} = \frac{5}{\sqrt{9 \sec^2 \theta + 4 \cos^2 \theta}} \]

### Step 4: Set the distance equal to the radius of the circle
The radius of the circle is \(\sqrt{3}\). Therefore, we set:
\[ \frac{5}{\sqrt{9 \sec^2 \theta + 4 \cos^2 \theta}} = \sqrt{3} \]
Squaring both sides gives:
\[ \frac{25}{9 \sec^2 \theta + 4 \cos^2 \theta} = 3 \]
This simplifies to:
\[ 25 = 3(9 \sec^2 \theta + 4 \cos^2 \theta) \]
Expanding this, we have:
\[ 25 = 27 \sec^2 \theta + 12 \cos^2 \theta \]

### Step 5: Substitute \(\sec^2 \theta\) in terms of \(\tan^2 \theta\)
We know that \(\sec^2 \theta = 1 + \tan^2 \theta\). Thus:
\[ 27(1 + \tan^2 \theta) + 12 \cos^2 \theta = 25 \]
This leads to:
\[ 27 + 27 \tan^2 \theta + 12 \cos^2 \theta = 25 \]
Rearranging gives:
\[ 27 \tan^2 \theta + 12 \cos^2 \theta = -2 \]
Since \(\tan^2 \theta\) and \(\cos^2 \theta\) are always non-negative, the left-hand side cannot be negative. Therefore, there are no values of \(\theta\) that satisfy this equation.

### Conclusion
Since there are no valid values of \(\theta\) that satisfy the conditions derived, we conclude that there are **zero tangents** to the circle \(x^2 + y^2 = 3\) that are normal to the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\).

### Final Answer
The number of tangents to the circle that are normal to the ellipse is **0**.

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