Prove that angle between one of the lines given by ax2-2hxy-by2-0 and one of the lines ax2-2hxy-by2- x2-y2-0 is equal to the angle between the other two lines of the system-

The two pairs have the same bisectors. x^2 y^2(a+λ) (b+λ)=xyh #160;or x^2 y^2a b=xyh and these are also the bisectors of 1 st pair. ...

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Standard XII

Mathematics

Angle between Two Lines

Prove that an...

Question

Prove that angle between one of the lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ and one of the lines $a x^{2} + 2 h x y + b y^{2} + λ (x^{2} + y^{2}) = 0$ is equal to the angle between the other two lines of the system.

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Solution

The two pairs have the same bisectors.

$\frac{x^{2} - y^{2}}{(a + λ) - (b + λ)} = \frac{x y}{h}$

or $\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h}$

and these are also the bisectors of $1$ st pair.

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Prove that angle between one of the lines given by ax2+2hxy+by2=0 and one of the lines ax2+2hxy+by2+λ(x2+y2)=0 is equal to the angle between the other two lines of the system.

The two pairs have the same bisectors.x2−y2(a+λ)−(b+λ)=xyh or x2−y2a−b=xyhand these are also the bisectors of 1st pair.

Prove that angle between one of the lines given by $a x^{2} + 2 h x y + b y^{2} = 0$ and one of the lines $a x^{2} + 2 h x y + b y^{2} + λ (x^{2} + y^{2}) = 0$ is equal to the angle between the other two lines of the system.

The two pairs have the same bisectors.

$\frac{x^{2} - y^{2}}{(a + λ) - (b + λ)} = \frac{x y}{h}$

or $\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h}$

and these are also the bisectors of $1$ st pair.