Question

let S=(1,2) andS'=(5,5)be the foci of a hyperbola.for any point P on the curve，it is given that|S'P-SP|=3.ife' is the eccentricity of the conjugate hyperbola then e' is =​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{e' = \frac{5}{4}}$

☆ɢɪᴠᴇɴ☆

• $S(1, 2), S'(5, 5)$ are foci of a hyperbola.

• $\left| S'P - SP \right| = 3$ for any point P on the hyperbola.

★ᴛᴏ ғɪɴᴅ★

• Eccentricity, e' of the conjugate hyperbola.

᯽ғᴏʀᴍᴜʟᴀ᯽

• Eccentricities of a hyperbola and its conjugate are related by

$\:\:\:\:\:\:\:\:\:\: \frac{1}{e^2} + \frac{1}{{e'}^2} = 1$

⍟ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ⍟

• It is clear that axes of the hyperbola are not x-axis and y-axis from the given data.

• Let us consider a new co-ordinate system in which X-axis and Y-axis coincides with the axes of the hyperbola. We can represent the hyperbola in this system as,

$\:\:\:\:\:\:\:\: \frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$

• Let O be the origin of the new co-ordinate system and A, A' be the vertices of the hyperbola.

Note:- Changing co-ordinate system does not affect distance between two points.

Now,

$\to \left| S'P - SP \right| = 3$

Since A is on the parabola,

$\to \left| S'A - SA \right| = 3$

$\to S'A - SA = 3$

$\to S'O + OA - (SO - OA) = 3$

$\to ae + a - (ae - a) = 3$

$\to 2a = 3.... (1)$

Also,

$\to SS' = 2ae$

$\to \sqrt{(5-1)^2 + (5-2)^2} = 2ae$

$\to 2ae = \sqrt{4^2 + 3^2}$

$\to 2ae = \sqrt{25}$

$\to 2ae = 5.....(2)$

Dividing (2) by (1), we get

$\to \boxed{\bf{ e = \frac{5}{3}}}$

Now, using the relation between e and e',

$\:\:\:\:\: \frac{1}{e^2} + \frac{1}{{e'}^2} = 1$

$\to \left(\frac{3}{5}\right)^2 + \frac{1}{{e'}^2} = 1$

$\to \frac{1}{{e'}^2} = 1 - \frac{9}{25}$

$\to \frac{1}{{e'}^2} = \frac{16}{25}$

$\to \frac{1}{e'} = \frac{4}{5}$

$\to \boxed{\bf{e'= \frac{5}{4}}}$