Question

If equation (10x – 5)² + (10y – 4)² = l² (3x + 4y – 1)² represents a hyperbola, then –

• A. –2 <l< 2
• B. l> 2
• C. l< –2 or l> 2
• D. 0 <l< 2

Answer 1

Answer: (C)

l< –2 or l> 2

Answer 2

Given equation of hyperbola

(10x – 5)² + (10y – 4)² = l²(3x + 4y – 1)²

can be rewritten as

$\frac{\sqrt{\left( x - \frac{1}{2} \right)^{2} + \left( y - \frac{2}{5} \right)^{2}}}{\left| \frac{3x + 4y - 1}{5} \right|}$= $\left| \frac{\lambda}{2} \right|$

This is of the form of $\frac{PS}{PM}$ = e

Where P is any point on the hyperbola and S is a focus and M is the point of directrix.

Here $\left| \frac{\lambda}{2} \right|$ > 1 Ž | l | > 2 [Q e > 1]

Ž l < –2 or l > 2.