$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represent two parallel lines if

h^{2} = a b, f^{2} > b c

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h^{2} = a b, h f = g b, f^{2} < b c

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h^{2} = a b, h f = g b, f^{2} > b c

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None of these

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Solution

The correct option is B $h^{2} = a b, h f = g b, f^{2} > b c$
It represent equation of pair of straight lines.
Now we know pair of lines are parallel if $\frac{h}{a} = \frac{b}{h} \to h^{2} = a b$
But this line can be coincident also so to be parallel,
$\frac{h}{b} = \frac{g}{f} \to h f = g b$
Now distance between two parallel lines is,
$∣ ∣ ∣ ∣ 2 \sqrt{\frac{f^{2} - b c}{b (a + b)}} ∣ ∣ ∣ ∣$
Means $f^{2} > b c$
Hence $(C)$

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Similar questions

Q. Statement A: The point of intersection of the lines represented by

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

(\frac{b g - h f}{h^{2} - a b}, \frac{a f - g h}{h^{2} - a b})

Statement B: The point of intersection of

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

(\sqrt{(\frac{f^{2} - b c}{h^{2} - a b})}, \sqrt{(\frac{g^{2} - a c}{h^{2} - a b})})

Q. If

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represents a pair of parallel lines, then

\sqrt{\frac{g^{2} - a c}{f^{2} - b c}} =

The square of distance between the point of intersection of the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ and origin, is

Q. If the equation

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then prove that the square of the distance of their point of intersection from the origin is

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Q. Show that the equation

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represents two parallel lines, if

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h^{2} = a b

and distance betwen these parallel lines is

2 \sqrt{\frac{g^{2} - a c}{a (a + b)}}

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ax2+2hxy+by2+2gx+2fy+c=0 represent two parallel lines if

$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ represent two parallel lines if