IIT-JEE 2008 · Paper 1 · Mathematics Q4

MathematicsSingle correct+3 / -1Question 4 of 69

Let $a$ and $b$ be non-zero real numbers. Then, the equation
$(ax^2+by^2+c)(x^2-5xy+6y^2)=0$
represents

four straight lines, when

c=0

and

a,b

are of the same sign

two straight lines and a circle, when

a=b

, and

c

is of sign opposite to that of

a

two straight lines and a hyperbola, when

a

and

b

are of the same sign and

c

is of sign opposite to that of

a

a circle and an ellipse, when

a

and

b

are of the same sign and

c

is of sign opposite to that of

a

Answer and worked solution

Official answer: B

Second factor. $x^2-5xy+6y^2=(x-2y)(x-3y)$ : two straight lines through the origin.

First factor. With $a=b$ , $ax^2+ay^2+c=0$ is $x^2+y^2=-\frac ca$ , a real circle when $c$ and $a$ have opposite signs.

The other options fail: (A) gives only the origin from the first factor; (C) gives an ellipse, not a hyperbola; (D) ignores the two lines.

(B)

IIT-JEE 2008 · Paper 1, Mathematics question 4