Question

Tangents are drawn from point $ (1, 1) $ to the ellipse $ x^2 + y^2 + 10x + 8y - 23 = 0 $. If $ m_1, m_2 $ (with $ m_1>m_2 $) are the slopes of these tangents, then with respect to the given ellipse, the point $ P(m_1, m_2) $ lies:

• A. lies inside the ellipse \( S = 0 \)
• B. lies outside the ellipse \( S = 0 \)
• C. lies on the ellipse \( S = 0 \)
• D. is the centre of the ellipse \( S = 0 \)

Hint:

The tangents from an external point to an ellipse always lie outside the ellipse, but the point formed by their slopes lies inside the ellipse when considered as a point in slope space.

Answer 1

The Correct Option is A

The given equation of the ellipse is \( x^2 + y^2 + 10x + 8y - 23 = 0 \). Let's first rewrite it in standard form by completing the square. \[ x^2 + 10x + y^2 + 8y = 23 \] Complete the square for both \( x \) and \( y \): \[ (x+5)^2 - 25 + (y+4)^2 - 16 = 23 \] \[ (x+5)^2 + (y+4)^2 = 64 \] This represents a circle with center \( (-5, -4) \) and radius 8. Now, the tangents drawn from the external point \( (1, 1) \) to this ellipse will form two tangents.
Since the tangents from a point outside an ellipse are always real and distinct, we will find the point \( P(m_1, m_2) \) where \( m_1 \) and \( m_2 \) are the slopes of these tangents.
The tangents to the ellipse form a curve, and the point \( P(m_1, m_2) \) lies inside the ellipse, as the tangents from the external point meet the curve inside the ellipse.
Thus, the correct answer is \( \boxed{\text{lies inside the ellipse}} \).