Question

For the hyperbola $16x^2 - 25y^2 - 32x + 50y - 409 = 0$. Which of the following is/are true?

• A. foci are $(\sqrt{41}+1, 1) \& (-\sqrt{41}+1, 1)$
• B. eccentricity is $\frac{\sqrt{41}}{5}$
• C. one of directrices $x = \frac{-25}{\sqrt{41}} + 1$
• D. length of latus rectum is $\frac{16}{5}$

Answer 1

Answer: (A), (B), (C)

Statements 1, 2, and 3 are true.

Answer 2

To determine which statements are true, we need to analyze the given hyperbola equation: $16x^2 - 25y^2 - 32x + 50y - 409 = 0$.

1. Complete the squares:

   Rearrange the equation: $16x^2 - 32x - 25y^2 + 50y = 409$

   Factor: $16(x^2 - 2x) - 25(y^2 - 2y) = 409$

   Complete the squares: $16[(x-1)^2 - 1] - 25[(y-1)^2 - 1] = 409$ $16(x-1)^2 - 16 - 25(y-1)^2 + 25 = 409$ $16(x-1)^2 - 25(y-1)^2 = 400$

   Divide by 400: $\frac{(x-1)^2}{25} - \frac{(y-1)^2}{16} = 1$

   This is a hyperbola centered at $(1, 1)$ with $a^2 = 25$ and $b^2 = 16$, so $a = 5$ and $b = 4$.

2. Foci:

   $c^2 = a^2 + b^2 = 25 + 16 = 41$, so $c = \sqrt{41}$. The foci are at $(h \pm c, k)$, which are $(1 \pm \sqrt{41}, 1)$. Statement 1 is true.

3. Eccentricity:

   $e = \frac{c}{a} = \frac{\sqrt{41}}{5}$. Statement 2 is true.

4. Directrices:

   The directrices are given by $x = h \pm \frac{a}{e}$. Since $\frac{a}{e} = \frac{5}{\frac{\sqrt{41}}{5}} = \frac{25}{\sqrt{41}}$, the directrices are $x = 1 \pm \frac{25}{\sqrt{41}}$. One of the directrices is $x = 1 - \frac{25}{\sqrt{41}}$. Statement 3 is true.

5. Length of Latus Rectum:

   The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 \cdot 16}{5} = \frac{32}{5}$. Statement 4 is false.

Therefore, statements 1, 2, and 3 are true.