If one of the lines of $a x^{2} + 2 h x y + b y^{2} = 0$ bisects the angle between the axes in the first quadrant, then

h^{2} - a b = 0

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h^{2} + a b = 0

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(a + b)^{2} = h^{2}

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(a + b)^{2} = 4 h^{2}

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Solution

The correct option is D $(a + b)^{2} = 4 h^{2}$
Since, one of the lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ bisects the angle between the axes in the first quadrant, therefore is equation is $y = x$ .
So, $y = x$ satisfies $a x^{2} + 2 h x y + b y^{2} = 0$ .
$a x^{2} + 2 h x^{2} + b x^{2} = 0$
$\Rightarrow (a + b) = - 2 h$
$\Rightarrow (a + b)^{2} = 4 h^{2}$ .

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Similar questions

Q. If one of the lines denoted by the line pair

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

bisects the angle between coordinate axes, then prove that

(a - b)^{2} = 4 h^{2}

Q. If

(a + b)^{2} = 4 h^{2}

, prove that one of the lines given by the equation

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

will bisect the angle between the co-ordinate axes.

If the equation $a x^{2} + 2 h x y + b y^{2} = 0$ has one line as the bisector of angle between the coordinate axes, then

If the pair of straight lines given by $A x^{2} + 2 H x y + B y^{2} = 0, (H^{2} > A B)$ forms an equilateral triangle with line ax + by + c = 0, then (A + 3B)(3A + B) is

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})

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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})})

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If one of the lines of ax2+2hxy+by2=0 bisects the angle between the axes in the first quadrant, then

The correct option is D (a+b)2=4h2Since, one of the lines represented by ax2+2hxy+by2=0 bisects the angle between the axes in the first quadrant, therefore is equation is y=x.So, y=x satisfies ax2+2hxy+by2=0.ax2+2hx2+bx2=0⇒(a+b)=−2h⇒(a+b)2=4h2.

If one of the lines of $a x^{2} + 2 h x y + b y^{2} = 0$ bisects the angle between the axes in the first quadrant, then