If the tangents drawn from P to the ellipse `(x^(2))/(25)+(y^(2))/(15)=1` are such that product of the slopes is equal to k then, find k such tha the locus of P is ellipse.

Answer

To solve the problem, we need to find the value of \( k \) such that the locus of point \( P \) from which tangents are drawn to the ellipse

\[ \frac{x^2}{25} + \frac{y^2}{15} = 1 \]

is also an ellipse.

### Step-by-Step Solution:

1. **Identify the Ellipse Parameters**:
The given ellipse can be expressed in the standard form where \( a^2 = 25 \) and \( b^2 = 15 \). Thus, \( a = 5 \) and \( b = \sqrt{15} \).

2. **Equation of the Tangent**:
The equation of the tangent to the ellipse at point \( (x_1, y_1) \) is given by:
\[ \frac{xx_1}{25} + \frac{yy_1}{15} = 1 \]
For a point \( P(\lambda, \mu) \), the tangent equation can be rewritten as:
\[ y = mx + \sqrt{25m^2 + 15} \]
and
\[ y = mx - \sqrt{25m^2 + 15} \]

3. **Substituting Point \( P \)**:
The point \( P(\lambda, \mu) \) lies on the tangent, so substituting \( x = \lambda \) and \( y = \mu \) gives:
\[ \mu = m\lambda \pm \sqrt{25m^2 + 15} \]

4. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ \mu - m\lambda = \pm \sqrt{25m^2 + 15} \]
Squaring both sides results in:
\[ (\mu - m\lambda)^2 = 25m^2 + 15 \]

5. **Expanding and Rearranging**:
Expanding the left side:
\[ \mu^2 - 2\mu m\lambda + m^2\lambda^2 = 25m^2 + 15 \]
Rearranging gives:
\[ m^2(\lambda^2 - 25) - 2\mu m\lambda + (\mu^2 - 15) = 0 \]

6. **Condition for Real Slopes**:
For \( m \) to have real values, the discriminant of this quadratic in \( m \) must be non-negative:
\[ D = (-2\mu \lambda)^2 - 4(\lambda^2 - 25)(\mu^2 - 15) \geq 0 \]

7. **Finding the Product of Slopes**:
The product of the slopes \( m_1 \) and \( m_2 \) can be found using the relation:
\[ k = \frac{c}{a} \]
where \( c = \mu^2 - 15 \) and \( a = \lambda^2 - 25 \). Thus:
\[ k = \frac{\mu^2 - 15}{\lambda^2 - 25} \]

8. **Condition for Locus to be an Ellipse**:
For the locus of \( P \) to be an ellipse, we require:
\[ k > 0 \quad \text{and} \quad \lambda^2 - 25 \neq 0 \]
This leads to the conditions on \( k \):
\[ k \neq -1 \quad \text{and} \quad k < 0 \]

9. **Conclusion**:
The final result for \( k \) such that the locus of \( P \) is an ellipse is:
\[ k \in (-\infty, -1) \cup (-1, 0) \]