Q.

The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and x=±43, respectively. Let the line y–3x+3=0 touch this hyperbola at (x0, y0). If m is the product of the focal distances of the point (x0, y0), then 4e2+m is equal to __________.          [2024]


Ans.

(*)

Equation of tangent to hyperbola is given by y=mx±a2m2–b2

⇒ m=3

and  a2m2–b2=3

⇒ 3a2–b2=3          ... (i)

Now, length of latus rectum of hyperbola = 9

⇒ 2b2a=9

⇒ b2=9a2          ... (ii)

From equation (i) and (ii), we get

3a2–9a2=3

⇒ 6a2–9a–6=0 ⇒ 2a2–3a–2=0

⇒ 2a2–4a+a–2=0 ⇒ (2a+1)(a–2)=0

⇒ a=2, –12 (ignore) ⇒ a=2  and  b=3

Equation of hyperbola is x24–y29=1

Solving for tangent y=3x–3, we get

x24–3x2+3–6x9=1

⇒ 9x2–12x2–12+24x=36

⇒ 3x2–24x+48=0

⇒ x2–8x+16=0 ⇒ (x–4)2=0

⇒ x=4  and  y=33, e=1+94=132

So, (x0,y0)≡(4,33) is the point of contact.

Focus of hyperbola is (±13, 0)

∴  PF1·PF2

    =((4–13)2+(33)2)((4+13)2+(33)2)

    =2304=48=m

∴  4e2+m=4×13/4+48=61

Note: Here equation of directrix should be x=±413, but in given question it is given x=±43 which is wrong because eccentricity should be greater than 1, So, this question is bonus.