Number of perpendicular tangents that can be drawn on the ellipse `(x^(2))/(16)+(y^(2))/(25)=1` from point (6, 7) is

Answer

To find the number of perpendicular tangents that can be drawn from the point (6, 7) to the ellipse given by the equation \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), we can follow these steps:

### Step 1: Identify the ellipse parameters
The given ellipse can be rewritten in standard form:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a^2 = 16\) and \(b^2 = 25\). Thus, \(a = 4\) and \(b = 5\).

### Step 2: Calculate the distance from the center of the ellipse to the point
The center of the ellipse is at the origin (0, 0). We need to calculate the distance \(d\) from the center of the ellipse to the point (6, 7):
\[ d = \sqrt{(6 - 0)^2 + (7 - 0)^2} = \sqrt{36 + 49} = \sqrt{85} \]

### Step 3: Determine the semi-major and semi-minor axes
From the ellipse parameters, we have:
- Semi-major axis \(b = 5\)
- Semi-minor axis \(a = 4\)

### Step 4: Use the formula for the number of tangents
The number of tangents \(N\) from an external point to the ellipse can be determined using the condition:
\[ N = \begin{cases} 2 & \text{if } d^2 > a^2 + b^2 \\ 1 & \text{if } d^2 = a^2 + b^2 \\ 0 & \text{if } d^2 < a^2 + b^2 \end{cases} \]

### Step 5: Calculate \(d^2\) and \(a^2 + b^2\)
First, calculate \(d^2\):
\[ d^2 = 85 \]
Now calculate \(a^2 + b^2\):
\[ a^2 + b^2 = 16 + 25 = 41 \]

### Step 6: Compare \(d^2\) with \(a^2 + b^2\)
Since \(d^2 = 85\) and \(a^2 + b^2 = 41\):
\[ 85 > 41 \]
Thus, according to our earlier condition, we have \(N = 2\).

### Conclusion
The number of perpendicular tangents that can be drawn from the point (6, 7) to the ellipse is **2**.

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