The equation frac{x^2}{a} + frac{xy}{h} + fra | vedclass

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(B) The given equation is $\frac{x^2}{a} + \frac{xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$,we get $bhx^2 + abxy + ahy^2 = 0$. For the lines to

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$4h^2 = ab$

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(B) The given equation is $\frac{x^2}{a} + \frac{xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$,we get $bhx^2 + abxy + ahy^2 = 0$. For the lines to be coincident,the equation must be of the form $(px + qy)^2 = 0$,which is $p^2x^2 + 2pqxy + q^2y^2 = 0$. Comparing the coefficients of the two equations: $\frac{bh}{p^2} = \frac{ab}{2pq} = \frac{ah}{q^2} = k$ (constant). From $\frac{bh}{p^2} = \frac{ah}{q^2}$,we get $\frac{b}{p^2} = \frac{a}{q^2}$ $\Rightarrow \frac{q^2}{p^2} = \frac{a}{b}$ $\Rightarrow \frac{q}{p} = \sqrt{\frac{a}{b}}$. From $\frac{ab}{2pq} = \frac{bh}{p^2}$,we get $\frac{a}{2q} = \frac{h}{p} \Rightarrow \frac{p}{q} = \frac{2h}{a}$. Equating the ratios: $\sqrt{\frac{b}{a}} = \frac{2h}{a}$. Squaring both sides: $\frac{b}{a} = \frac{4h^2}{a^2}$ $\Rightarrow b = \frac{4h^2}{a}$ $\Rightarrow ab = 4h^2$. Thus,the condition for coincident lines is $4h^2 = ab$.

The equation $\frac{x^2}{a} + \frac{xy}{h} + \frac{y^2}{b} = 0$ $(a \neq 0, h \neq 0, b \neq 0)$ represents two coincident lines if:

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