Question

Let $P$ be any point on the ellipse $4(x+2)^{2}+9(y-4)^{2}=144$. If $F_{1}$ and $F_{2}$ are the foci of the ellipse, then $F_{1}P+F_{2}P=$

• A. 8
• B. 12
• C. 16
• D. 6
• E. 10

Hint:

The definition of an ellipse is the locus of points where the sum of distances to two fixed points (foci) is constant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question uses the fundamental definition of an ellipse. An ellipse is the set of all points P in a plane such that the sum of the distances from P to two fixed points (the foci, \( F_1 \) and \( F_2 \)) is a constant. This constant sum, \( F_1P + F_2P \), is equal to the length of the major axis of the ellipse, which is \( 2a \).
Step 2: Key Formula or Approach:
1. The definition of an ellipse: \( F_1P + F_2P = 2a \).
2. The standard form of an ellipse's equation is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) (for a horizontal major axis) or \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) (for a vertical major axis), where \( a>b \).
Step 3: Detailed Explanation:
First, we must convert the given equation of the ellipse into its standard form to identify the value of 'a'.
The given equation is:
\[ 4(x+2)^2 + 9(y-4)^2 = 144 \] To get 1 on the right-hand side, divide the entire equation by 144:
\[ \frac{4(x+2)^2}{144} + \frac{9(y-4)^2}{144} = \frac{144}{144} \] Simplify the fractions:
\[ \frac{(x+2)^2}{36} + \frac{(y-4)^2}{16} = 1 \] Now, compare this with the standard form. We can identify \( a^2 \) and \( b^2 \). Since \( 36>16 \), the major axis is horizontal.
\[ a^2 = 36 \] \[ b^2 = 16 \] From \( a^2 = 36 \), we find the semi-major axis length:
\[ a = \sqrt{36} = 6 \] According to the definition of the ellipse, the sum of the distances from any point on the ellipse to the foci is equal to the length of the major axis, \( 2a \).
\[ F_1P + F_2P = 2a = 2(6) = 12 \] Step 4: Final Answer:
The value of \( F_1P + F_2P \) is 12.