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Question

If $$(a+b)^2=4h^2$$, prove that one of the lines given by the equation $$ax^2+2hxy+by^2=0$$ will bisect the angle between the co-ordinate axes.


Solution

Solution:-
The equation of the bisectors of the angles between the coordinate axes are $$y = x$$ and $$y = -x$$.

Let $$y = x$$ belong to $$a{x}^{2} + 2hxy + b{y}^{2} = 0$$

$$\therefore a{x}^{2} + 2h {x}^{2} + b{x}^{2} = 0$$

$$\Rightarrow a + b = -2h ..... \left( 1 \right)$$

If $$y = -x$$ belongs to $$a{x}^{2} + 2hxy + b{y}^{2} = 0$$.

$$\therefore a{x}^{2} - 2h{x}^{2} + b{x}^{2} = 0$$

$$\Rightarrow a + b = 2h ..... \left( 2 \right)$$

From $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have

$$\left( a + b \right) = \pm 2h$$

Squaring both sides, we get

$${\left( a + b \right)}^{2} = 4{h}^{2}$$

Hence proved.

Maths

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