If the tangent at the point (h,k) to `(x^2)/(a^2)+(y^2)/(b^2)=1` meets the circle `x^2+y^2=a^2` at `(x_1,y_1)" and "(x_2,y_2)`, then `y_1,k,y_2` are in

Options

A. A.P
B. G.P
C. H.P
D. None of these

Answer

To solve the problem, we need to find the relationship between \( y_1, k, \) and \( y_2 \) given that the tangent at the point \( (h, k) \) to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) intersects the circle \( x^2 + y^2 = a^2 \) at points \( (x_1, y_1) \) and \( (x_2, y_2) \).

### Step 1: Write the equation of the tangent line
The equation of the tangent to the ellipse at the point \( (h, k) \) is given by:
\[ \frac{hx}{a^2} + \frac{ky}{b^2} = 1 \]

### Step 2: Substitute \( y \) in the circle's equation
The equation of the circle is:
\[ x^2 + y^2 = a^2 \]
We can express \( y \) from the tangent equation:
\[ ky = b^2(1 - \frac{hx}{a^2}) \implies y = \frac{b^2}{k}(1 - \frac{hx}{a^2}) \]
Now substitute this expression for \( y \) into the circle's equation:
\[ x^2 + \left(\frac{b^2}{k}\left(1 - \frac{hx}{a^2}\right)\right)^2 = a^2 \]

### Step 3: Expand and rearrange the equation
Expanding the equation gives:
\[ x^2 + \frac{b^4}{k^2}\left(1 - \frac{2hx}{a^2} + \frac{h^2x^2}{a^4}\right) = a^2 \]
\[ x^2 + \frac{b^4}{k^2} - \frac{2b^4h}{k^2a^2}x + \frac{b^4h^2}{k^2a^4}x^2 = a^2 \]
Combining like terms:
\[ \left(1 + \frac{b^4h^2}{k^2a^4}\right)x^2 - \frac{2b^4h}{k^2a^2}x + \left(\frac{b^4}{k^2} - a^2\right) = 0 \]

### Step 4: Identify coefficients for the quadratic equation
Let:
- \( A = 1 + \frac{b^4h^2}{k^2a^4} \)
- \( B = -\frac{2b^4h}{k^2a^2} \)
- \( C = \frac{b^4}{k^2} - a^2 \)

### Step 5: Use the properties of the roots
From the quadratic equation \( Ax^2 + Bx + C = 0 \), we know:
- The sum of the roots \( y_1 + y_2 = -\frac{B}{A} \)
- The product of the roots \( y_1y_2 = \frac{C}{A} \)

### Step 6: Express the sum and product in terms of \( k \)
Substituting \( B \) and \( C \):
\[ y_1 + y_2 = \frac{2b^4h}{k^2a^2 \left(1 + \frac{b^4h^2}{k^2a^4}\right)} \]
\[ y_1y_2 = \frac{\left(\frac{b^4}{k^2} - a^2\right)}{1 + \frac{b^4h^2}{k^2a^4}} \]

### Step 7: Establish the relationship
To show that \( y_1, k, y_2 \) are in Harmonic Progression (HP), we need to show:
\[ \frac{1}{y_1} + \frac{1}{y_2} = \frac{2}{k} \]
This can be derived from the expressions for \( y_1 + y_2 \) and \( y_1y_2 \).

### Conclusion
Thus, we conclude that \( y_1, k, y_2 \) are in Harmonic Progression.