If tangents are drawn to the ellipse `2x^(2) + 3y^(2) =6`, then the locus of the mid-point of the intercept made by the tangents between the co-ordinate axes is

Options

A. `(1)/(4x^(2)) + (1)/(2y^(2)) =1`
B. `(3)/(4x^(2)) + (1)/(2y^(2)) =1`
C. `(1)/(x^(2)) + (1)/(y^(2)) =1`
D. `(1)/(x^(2)) + (3)/(4y^(2)) =1`

Answer

To find the locus of the midpoint of the intercept made by the tangents to the ellipse \(2x^2 + 3y^2 = 6\) between the coordinate axes, we will follow these steps:

### Step 1: Rewrite the equation of the ellipse
We start with the given equation of the ellipse:
\[ 2x^2 + 3y^2 = 6 \]
Dividing through by 6 gives:
\[ \frac{x^2}{3} + \frac{y^2}{2} = 1 \]

### Step 2: Identify the semi-major and semi-minor axes
From the standard form of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we identify:
- \(a^2 = 3 \Rightarrow a = \sqrt{3}\)
- \(b^2 = 2 \Rightarrow b = \sqrt{2}\)

### Step 3: Parametric equations of the ellipse
The parametric equations for the ellipse are given by:
\[ x = a \cos \theta = \sqrt{3} \cos \theta, \quad y = b \sin \theta = \sqrt{2} \sin \theta \]

### Step 4: Equation of the tangent line
The equation of the tangent to the ellipse at the point \((\sqrt{3} \cos \theta, \sqrt{2} \sin \theta)\) can be derived using the formula:
\[ \frac{xx_1}{3} + \frac{yy_1}{2} = 1 \]
Substituting \(x_1 = \sqrt{3} \cos \theta\) and \(y_1 = \sqrt{2} \sin \theta\):
\[ \frac{x \sqrt{3} \cos \theta}{3} + \frac{y \sqrt{2} \sin \theta}{2} = 1 \]

### Step 5: Intercepts on the axes
To find the x-intercept (set \(y = 0\)):
\[ \frac{x \sqrt{3} \cos \theta}{3} = 1 \Rightarrow x = \frac{3}{\sqrt{3} \cos \theta} = \frac{\sqrt{3}}{\cos \theta} \]
To find the y-intercept (set \(x = 0\)):
\[ \frac{y \sqrt{2} \sin \theta}{2} = 1 \Rightarrow y = \frac{2}{\sqrt{2} \sin \theta} = \frac{\sqrt{2}}{\sin \theta} \]

### Step 6: Midpoint of the intercepts
The intercepts are \(\left(\frac{\sqrt{3}}{\cos \theta}, 0\right)\) and \(\left(0, \frac{\sqrt{2}}{\sin \theta}\right)\). The midpoint \((h, k)\) of these intercepts is given by:
\[ h = \frac{1}{2} \left(\frac{\sqrt{3}}{\cos \theta} + 0\right) = \frac{\sqrt{3}}{2 \cos \theta} \]
\[ k = \frac{1}{2} \left(0 + \frac{\sqrt{2}}{\sin \theta}\right) = \frac{\sqrt{2}}{2 \sin \theta} \]

### Step 7: Relate \(h\) and \(k\)
From the expressions for \(h\) and \(k\):
\[ \sin \theta = \frac{\sqrt{2}}{2k} \quad \text{and} \quad \cos \theta = \frac{\sqrt{3}}{2h} \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \left(\frac{\sqrt{2}}{2k}\right)^2 + \left(\frac{\sqrt{3}}{2h}\right)^2 = 1 \]
This simplifies to:
\[ \frac{2}{4k^2} + \frac{3}{4h^2} = 1 \]
Multiplying through by 4 gives:
\[ \frac{2}{k^2} + \frac{3}{h^2} = 4 \]

### Step 8: Rearranging to find the locus
Rearranging gives:
\[ \frac{3}{h^2} = 4 - \frac{2}{k^2} \]
This can be rewritten as:
\[ \frac{3}{h^2} + \frac{2}{k^2} = 4 \]
Replacing \(h\) and \(k\) with \(x\) and \(y\):
\[ \frac{3}{4x^2} + \frac{1}{2y^2} = 1 \]

### Final Answer
Thus, the locus of the midpoint of the intercepts made by the tangents to the ellipse is:
\[ \frac{3}{4x^2} + \frac{1}{2y^2} = 1 \]