Number of points on the ellipse `(x^(2))/(25) + (y^(2))/(16) =1` from which pair of perpendicular tangents are drawn to the ellipse `(x^(2))/(16) + (y^(2))/(9) =1` is

Options

A. 2
B. 3
C. 4
D. 1

Answer

To solve the problem of finding the number of points on the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) from which a pair of perpendicular tangents can be drawn to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\), we can follow these steps:

### Step 1: Identify the properties of the ellipses
The first ellipse is given by the equation:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
This means \(a_1^2 = 25\) and \(b_1^2 = 16\), where \(a_1 = 5\) and \(b_1 = 4\). The center of this ellipse is at the origin (0, 0).

The second ellipse is given by the equation:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
This means \(a_2^2 = 16\) and \(b_2^2 = 9\), where \(a_2 = 4\) and \(b_2 = 3\). The center of this ellipse is also at the origin (0, 0).

### Step 2: Use the condition for perpendicular tangents
For a point \((x_1, y_1)\) on the first ellipse, the condition for drawing perpendicular tangents to the second ellipse is given by the equation:
\[ \frac{x_1^2}{a_2^2} + \frac{y_1^2}{b_2^2} = 1 \]
This can be rewritten using the values of \(a_2\) and \(b_2\):
\[ \frac{x_1^2}{16} + \frac{y_1^2}{9} = 1 \]

### Step 3: Substitute the parametric equations of the first ellipse
The points on the first ellipse can be expressed in parametric form:
\[ x_1 = 5 \cos \theta, \quad y_1 = 4 \sin \theta \]
Substituting these into the perpendicular tangent condition gives:
\[ \frac{(5 \cos \theta)^2}{16} + \frac{(4 \sin \theta)^2}{9} = 1 \]
This simplifies to:
\[ \frac{25 \cos^2 \theta}{16} + \frac{16 \sin^2 \theta}{9} = 1 \]

### Step 4: Clear the denominators
To eliminate the fractions, multiply through by \(144\) (the least common multiple of \(16\) and \(9\)):
\[ 144 \left(\frac{25 \cos^2 \theta}{16} + \frac{16 \sin^2 \theta}{9}\right) = 144 \]
This results in:
\[ 225 \cos^2 \theta + 256 \sin^2 \theta = 144 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 225 \cos^2 \theta + 256 \sin^2 \theta - 144 = 0 \]

### Step 6: Solve the quadratic in \(\cos^2 \theta\)
Let \(u = \cos^2 \theta\). Then, \(\sin^2 \theta = 1 - u\). Substitute this into the equation:
\[ 225u + 256(1 - u) - 144 = 0 \]
This simplifies to:
\[ 225u + 256 - 256u - 144 = 0 \]
\[ -31u + 112 = 0 \]
\[ u = \frac{112}{31} \]

### Step 7: Analyze the solution
Since \(u = \cos^2 \theta\) must be in the interval \([0, 1]\), we check if \(\frac{112}{31} > 1\). Since it is greater than 1, there are no valid solutions for \(u\).

### Conclusion
Thus, there are no points on the first ellipse from which a pair of perpendicular tangents can be drawn to the second ellipse.

### Final Answer
The number of points on the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) from which a pair of perpendicular tangents can be drawn to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) is **0**.