If a hyperbola has one focus at the origin and its eccentricity is -2- One of the directrices is x - y -1-0- Then the equations to its asymptotes areA- x-1-0- y-1-0B- x-1-0- y-1-0C- x-3-0- y-3-0D- x-2-0- y-2-0

The correct option is B Let P(x, y) be a point on the hyperbola, then SP^2 = e^2 PM^2⇒ x^2 + y^2 = 2(x+y+1/√(2))^2 ⇒ 2xy + 2x + 2y + 1 = 0 ∴Equation of its pa ...

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Standard XII

Mathematics

Conjugate Hyperbola

If a hyperbol...

Question

If a hyperbola has one focus at the origin and its eccentricity is $\sqrt{2}$ . One of the directrices is $x + y + 1 = 0.$ Then the equations to its asymptotes are

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Solution

The correct option is B
Let P(x, y) be a point on the hyperbola, then $S P^{2} = e^{2} P M^{2} \Rightarrow x^{2} + y^{2} = 2 {(\frac{x + y + 1}{\sqrt{2}})}^{2}$
$\Rightarrow 2 x y + 2 x + 2 y + 1 = 0$
$∴$ Equation of its pair of asymptotes is $2 x y + 2 x + 2 y + k = 0$
Applying $Δ = 0$ , we get $k = 2$
$∴$ Equation of asymptotes is $x y + x + y + 1 = 0 \Rightarrow x + 1 = 0; y + 1 = 0$

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If a hyperbola has one focus at the origin and its eccentricity is √2 . One of the directrices is x+y+1=0. Then the equations to its asymptotes are

The correct option is B Let P(x, y) be a point on the hyperbola, then SP2=e2PM2⇒x2+y2=2(x+y+1√2)2 ⇒2xy+2x+2y+1=0 ∴Equation of its pair of asymptotes is 2xy+2x+2y+k=0 Applying Δ=0, we get k=2 ∴Equation of asymptotes is xy+x+y+1=0⇒x+1=0;y+1=0

If a hyperbola has one focus at the origin and its eccentricity is $\sqrt{2}$ . One of the directrices is $x + y + 1 = 0.$ Then the equations to its asymptotes are