The equation to the pair of lines passing through the point $(3, - 2)$ and perpendicular to the pair of lines $x^{2} - 2 x y - 8 y^{2} = 0$

8 x^{2} + 2 x y - y^{2} - 52 x + 2 y - 80 = 0

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8 x^{2} + 2 x y - y^{2} + 52 x + 2 y - 80 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 x^{2} - 2 x y - y^{2} - 52 x + 2 y + 80 = 0

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8 x^{2} - 2 x y - y^{2} + 52 x + 2 y + 80 = 0

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Solution

The correct option is C $8 x^{2} - 2 x y - y^{2} - 52 x + 2 y + 80 = 0$
$x^{2} - 2 x y - 8 y^{2} = 0$
Divides both sides by $x^{2}$
$\Rightarrow 1 - 2 (\frac{y}{x}) - 8 {(\frac{y}{x})}^{2} = 0$
$\Rightarrow 8 m^{2} + 2 m - 1 = 0 (\frac{y}{x} = m =$ slope $)$
$\Rightarrow 2 m^{2} + 4 m - 2 m - 1 = 0$
$\Rightarrow 4 m (2 m + 1) - 1 (2 m + 1) = 0$
$\Rightarrow (2 m + 1) (4 m - 1) = 0$
$∴ m = - \frac{1}{2}, \frac{1}{4}$
slopes of lines perpendicular to it
wrel be $2, - 4 (m_{1} \cdot m_{2} = - 1)$
E $q^{n}$ of poir of straight lines $p$ assing
through (3,-2)
$(2 x - y - 8) (4 x + y - 10) = 0$
$\Rightarrow 8 x^{2} + 2 x y - 20 x - 4 x y - y^{2}$
$+ 10 y - 32 x - 8 y + 80 = 0$
$8 x^{2} - 2 x y - y^{2} - 52 x + 2 y + 80 = 0$

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The equation to the pair of lines passing through the point (3,−2) and perpendicular to the pair of lines x2−2xy−8y2=0

The equation to the pair of lines passing through the point $(3, - 2)$ and perpendicular to the pair of lines $x^{2} - 2 x y - 8 y^{2} = 0$