Question

The straight lines joining the origin to the intersection points of the curves whose equations are ax² + 2hxy + by² + 2gx = 0

and a¢x² + 2h¢xy + b¢y² + 2g¢x = 0 are at right angles if-

• A. (a + b)g¢ = (a¢ + b¢)g
• B. $\left( \frac{1}{a} + \frac{1}{b} \right)g' = \left( \frac{1}{a'} + \frac{1}{b'} \right)g$
• C. (g¢ + h¢) (a¢ + b¢) = (g + h) (a¢ + b¢)
• D. $\frac{1}{g'} + \frac{1}{h'} = \frac{1}{g} + \frac{1}{h}$

Answer 1

Answer: (A)

(a + b)g¢ = (a¢ + b¢)g

Answer 2

Let the two curves be S₁ and S₂. Then (g¢ × S₁) – (g × S₂) gives

(ag¢ – a¢g)x² + 2(g¢h – gh¢)xy + (bg¢ – b¢g)y² = 0

Which is a homogeneous second degree equation and thus represents a pair of straight lines passing through the origin.

The two lines will be at right angles if

coeff. of x² + coeff. of y² = 0

i.e. (a + b)g¢ = (a¢ + b¢)g .