Question

If (x₁, y₁) is a point on the hyperbola$\frac{x^{2}}{64}$–$\frac{y^{2}}{36}$= 1 in the first quadrant whose distance from right focus is $\frac{9}{2}$then (x₁ + y₁) must be

• A. $\frac{29}{2}$
• B. $\frac{19}{2}$
• C. $\frac{31}{2}$
• D. –7

Answer 1

Answer: (A)

$\frac{29}{2}$

Answer 2

$\frac{x^{2}}{64}$–$\frac{y^{2}}{36}$= 1 …(1)

e =$\sqrt{1 + \frac{36}{64}}$=$\frac{5}{4}$

Now,

PS = e.PM = e(NQ)

= e (CN – CQ) = e(x₁ – a/e)

= ex₁ – a = $\frac{9}{2}$

Ž $\frac{5}{4}$x₁ – 8 =$\frac{9}{2}$Ž x₁ = 10

By (1) Ž y₁ = ±$\frac{9}{2}$ Q y > 0

\ y₁ = $\frac{9}{2}$

\ (x₁ + y₁) = 10 + $\frac{9}{2}$= $\frac{29}{2}$