`5sqrt3x+7y=70` is a tangent to the ellipse `x^2/49+y^2/25=1` at the point P. Find the coordinates of P and the equation of the normal at P .

Answer

To solve the problem step by step, we will find the coordinates of the point \( P \) where the line \( 5\sqrt{3}x + 7y = 70 \) is tangent to the ellipse \( \frac{x^2}{49} + \frac{y^2}{25} = 1 \), and then we will derive the equation of the normal at that point.

### Step 1: Identify the parameters of the ellipse
The given ellipse is represented as:
\[ \frac{x^2}{49} + \frac{y^2}{25} = 1 \]
From this, we can identify:
- \( a^2 = 49 \) so \( a = 7 \)
- \( b^2 = 25 \) so \( b = 5 \)

### Step 2: Parametric representation of the ellipse
The parametric equations for the ellipse can be expressed as:
\[ x = a \cos \theta = 7 \cos \theta \]
\[ y = b \sin \theta = 5 \sin \theta \]

### Step 3: Write the equation of the tangent line at point \( P \)
The equation of the tangent to the ellipse at point \( P(a \cos \theta, b \sin \theta)\) is given by:
\[ \frac{x \cdot a \cos \theta}{a^2} + \frac{y \cdot b \sin \theta}{b^2} = 1 \]
Substituting \( a = 7 \) and \( b = 5 \):
\[ \frac{x \cdot 7 \cos \theta}{49} + \frac{y \cdot 5 \sin \theta}{25} = 1 \]
This simplifies to:
\[ \frac{x \cos \theta}{7} + \frac{y \sin \theta}{5} = 1 \]

### Step 4: Compare the coefficients with the tangent line equation
We are given the tangent line:
\[ 5\sqrt{3}x + 7y = 70 \]
Rearranging gives:
\[ \frac{x \cdot 5\sqrt{3}}{70} + \frac{y \cdot 7}{70} = 1 \]
This implies:
\[ \cos \theta = \frac{5\sqrt{3}}{70} \cdot 7 = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} \]
\[ \sin \theta = \frac{7}{70} \cdot 5 = \frac{1}{2} \]

### Step 5: Find the coordinates of point \( P \)
Using the values of \( \cos \theta \) and \( \sin \theta \):
\[ x = 7 \cos \theta = 7 \cdot \frac{\sqrt{3}}{2} = \frac{7\sqrt{3}}{2} \]
\[ y = 5 \sin \theta = 5 \cdot \frac{1}{2} = \frac{5}{2} \]
Thus, the coordinates of point \( P \) are:
\[ P\left(\frac{7\sqrt{3}}{2}, \frac{5}{2}\right) \]

### Step 6: Find the equation of the normal at point \( P \)
The equation of the normal to the ellipse at point \( P(a \cos \theta, b \sin \theta)\) is given by:
\[ \frac{a^2}{x_1^2} (x - x_1) + \frac{b^2}{y_1^2} (y - y_1) = 0 \]
Substituting \( x_1 = \frac{7\sqrt{3}}{2} \) and \( y_1 = \frac{5}{2} \):
\[ \frac{49}{\left(\frac{7\sqrt{3}}{2}\right)^2} \left(x - \frac{7\sqrt{3}}{2}\right) + \frac{25}{\left(\frac{5}{2}\right)^2} \left(y - \frac{5}{2}\right) = 0 \]
Calculating \( \left(\frac{7\sqrt{3}}{2}\right)^2 = \frac{49 \cdot 3}{4} = \frac{147}{4} \) and \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \):
\[ \frac{49 \cdot 4}{147} (x - \frac{7\sqrt{3}}{2}) + \frac{25 \cdot 4}{25} (y - \frac{5}{2}) = 0 \]
This simplifies to:
\[ \frac{28}{3} (x - \frac{7\sqrt{3}}{2}) + 4 (y - \frac{5}{2}) = 0 \]

### Final Normal Equation
After simplifying, we find the normal equation:
\[ 28x - 4y - 28\sqrt{3} + 10 = 0 \]

### Summary of Results
- Coordinates of point \( P \): \( \left(\frac{7\sqrt{3}}{2}, \frac{5}{2}\right) \)
- Equation of the normal at point \( P \): \( 28x - 4y - 28\sqrt{3} + 10 = 0 \)