If `y=x+c` touches the ellipse `3x^(2)+4y^(2)=12` at the point P, then the value of the length OP (where O is the origin) is equal to

Options

A. `sqrt3` units
B. `sqrt7` units
C. `(5)/(sqrt7)` units
D. `sqrt((7)/(5))` units

Answer

To find the length OP where O is the origin and P is the point where the line \(y = x + c\) touches the ellipse \(3x^2 + 4y^2 = 12\), we can follow these steps:

### Step 1: Rewrite the ellipse equation
The given ellipse equation is:
\[ 3x^2 + 4y^2 = 12 \]
We can divide the entire equation by 12 to get it into standard form:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
This indicates that \(a^2 = 4\) and \(b^2 = 3\), thus \(a = 2\) and \(b = \sqrt{3}\).

### Step 2: Parametric coordinates of point P
Using the parametric form of the ellipse, we can express the coordinates of point P as:
\[ P = (a \cos \theta, b \sin \theta) = (2 \cos \theta, \sqrt{3} \sin \theta) \]

### Step 3: Equation of the tangent at point P
The equation of the tangent to the ellipse at point P in parametric form is given by:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]
Substituting \(a\) and \(b\):
\[ \frac{x \cos \theta}{2} + \frac{y \sin \theta}{\sqrt{3}} = 1 \]

### Step 4: Compare with the given line equation
The tangent line is given as \(y = x + c\). We can rearrange the tangent equation:
\[ y = -\frac{2 \sin \theta}{\sqrt{3} \cos \theta} x + \frac{\sqrt{3}}{\sin \theta} \]
The slope of this tangent line is \(-\frac{2 \sin \theta}{\sqrt{3} \cos \theta}\). For the line \(y = x + c\), the slope is 1. Therefore, we set:
\[ -\frac{2 \sin \theta}{\sqrt{3} \cos \theta} = 1 \]
This gives us:
\[ 2 \sin \theta = -\sqrt{3} \cos \theta \]
Taking the cotangent:
\[ \cot \theta = \frac{2}{\sqrt{3}} \implies \tan \theta = \frac{\sqrt{3}}{2} \]

### Step 5: Find the coordinates of point P
From the value of \(\tan \theta\), we can form a right triangle where the opposite side is \(\sqrt{3}\) and the adjacent side is \(2\). By the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(\sqrt{3})^2 + (2)^2} = \sqrt{3 + 4} = \sqrt{7} \]
Now, we can find \(\cos \theta\) and \(\sin \theta\):
\[ \cos \theta = \frac{2}{\sqrt{7}}, \quad \sin \theta = \frac{\sqrt{3}}{\sqrt{7}} \]

### Step 6: Substitute back to find coordinates of P
Substituting \(\cos \theta\) and \(\sin \theta\) into the coordinates of P:
\[ P = \left(2 \cdot \frac{2}{\sqrt{7}}, \sqrt{3} \cdot \frac{\sqrt{3}}{\sqrt{7}}\right) = \left(\frac{4}{\sqrt{7}}, \frac{3}{\sqrt{7}}\right) \]

### Step 7: Calculate the distance OP
The distance OP from the origin O(0,0) to point P is given by:
\[ OP = \sqrt{\left(\frac{4}{\sqrt{7}} - 0\right)^2 + \left(\frac{3}{\sqrt{7}} - 0\right)^2} \]
Calculating this:
\[ = \sqrt{\frac{16}{7} + \frac{9}{7}} = \sqrt{\frac{25}{7}} = \frac{5}{\sqrt{7}} \]

Thus, the length OP is:
\[ \boxed{\frac{5}{\sqrt{7}}} \]