Question

S' is the focus of the ellipse \( \frac{x^2}{25} + \frac{y^2}{b^2} = 1, (b<5) \) lying on the negative X-axis and \( P(\theta) \) is a point on this ellipse. If the distance between the foci of this ellipse is 8 and \( S'P = 7 \), then \( \theta \) is:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{2\pi}{3} \) \bigskip

Hint:

To solve ellipse-related problems, use the relationships between the semi-major axis \( a \), semi-minor axis \( b \), and the foci distance. Apply the parametric equations of the ellipse to find the angle \( \theta \) based on the given distances.

Answer 1

The Correct Option is B

We are given the equation of the ellipse: \[ \frac{x^2}{25} + \frac{y^2}{b^2} = 1 \] The foci of the ellipse lie along the X-axis, and the distance between the foci is given as 8. Step 1: Use the formula for the distance between the foci The distance between the foci of the ellipse is given by \( 2c \), where \( c = \sqrt{a^2 - b^2} \). Here, \( a = 5 \) (since \( a^2 = 25 \)), and we are told the distance between the foci is 8. Thus: \[ 2c = 8 \quad \Rightarrow \quad c = 4 \] Now we can find \( b \) using the equation \( c = \sqrt{a^2 - b^2} \): \[ 4 = \sqrt{25 - b^2} \] Squaring both sides: \[ 16 = 25 - b^2 \] Solving for \( b^2 \): \[ b^2 = 9 \quad \Rightarrow \quad b = 3 \] Step 2: Apply the information about point \( P \) We are also given that \( S'P = 7 \), where \( S' \) is the focus at \( (-4, 0) \), and \( P(\theta) \) is a point on the ellipse. We know that the distance from the point \( P \) to the focus \( S' \) is \( S'P = 7 \). Using the parametric equations for an ellipse, the coordinates of a point on the ellipse are given by: \[ x = a \cos \theta, \quad y = b \sin \theta \] For \( P(\theta) \), the distance \( S'P \) is given by: \[ S'P = \sqrt{(x + 4)^2 + y^2} \] Substitute \( x = 5 \cos \theta \) and \( y = 3 \sin \theta \): \[ S'P = \sqrt{(5 \cos \theta + 4)^2 + (3 \sin \theta)^2} \] We are told that \( S'P = 7 \): \[ 7 = \sqrt{(5 \cos \theta + 4)^2 + 9 \sin^2 \theta} \] Squaring both sides: \[ 49 = (5 \cos \theta + 4)^2 + 9 \sin^2 \theta \] Expanding the terms: \[ 49 = 25 \cos^2 \theta + 40 \cos \theta + 16 + 9 \sin^2 \theta \] Since \( \cos^2 \theta + \sin^2 \theta = 1 \), substitute this: \[ 49 = 25 \cos^2 \theta + 9 (1 - \cos^2 \theta) + 40 \cos \theta + 16 \] Simplify: \[ 49 = 25 \cos^2 \theta + 9 - 9 \cos^2 \theta + 40 \cos \theta + 16 \] \[ 49 = 16 \cos^2 \theta + 40 \cos \theta + 25 \] Rearrange: \[ 0 = 16 \cos^2 \theta + 40 \cos \theta - 24 \] Solve the quadratic equation for \( \cos \theta \) using the quadratic formula: \[ \cos \theta = \frac{-40 \pm \sqrt{40^2 - 4 \cdot 16 \cdot (-24)}}{2 \cdot 16} \] \[ \cos \theta = \frac{-40 \pm \sqrt{1600 + 1536}}{32} \] \[ \cos \theta = \frac{-40 \pm \sqrt{3136}}{32} \] \[ \cos \theta = \frac{-40 \pm 56}{32} \] Thus, \( \cos \theta = \frac{16}{32} = \frac{1}{2} \), so \( \theta = \frac{\pi}{3} \). Thus, the correct value of \( \theta \) is \( \frac{\pi}{3} \). \bigskip