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Angle between Two Lines

The line lx...

Question

The line $l x + m y + n = 0$ intersects the curve $a x^{2} + 2 h x y + b y^{2} = 1$ at $P$ and $Q$ . The circle with $P Q$ as diameter passes through the origin then $\frac{l^{2} + m^{2}}{n^{2}} =$

a + b

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(a + b)^{2}

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a^{2} + b^{2}

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a^{2} - b^{2}

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Solution

The correct option is C $a + b$
$l x + m y + x = 0$
$y = \frac{- x - l x}{m}$ ---1
$∴$ $a x^{2} + 2 h x y + b y^{2} = 1$
Put $y = \frac{- x - l x}{m}$ in the above equation
$a x^{2} + 2 h x (\frac{- h - l x}{m}) + b (\frac{h + l x}{m})^{2} = 1$
$a x^{2} - \frac{2 x h x}{m} - \frac{2 h l x^{2}}{m} + \frac{b x^{2}}{m^{2}} + \frac{b l^{2} x^{2}}{m^{2}} + \frac{2 b x l x}{m} = 1$
$(a - \frac{2 h l}{m} + \frac{b l^{2}}{m^{2}}) x^{2} + (\frac{2 b x l}{m} - \frac{2 x h}{m}) x + \frac{b x^{2}}{m^{2}} - 1 = 0$
$∴$ $x_{2} + x_{2} = \frac{\frac{2 x h - 2 b x l}{m}}{a - \frac{2 h l}{m} + \frac{b l^{2}}{m^{2}}}$

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Similar questions

l x + m y + n = 0

a x^{2} + 2 h x y + b y^{2} = 1

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\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = {(\frac{a^{2} + b^{2}}{n})}^{2}

l x + m y + n = 0

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l x + m y = n

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\frac{n^{2}}{(a^{2} - b^{2})^{2}} (\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}}) = k,

7 k =

l x + m y + n = 0

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}}

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