The equation of the hyperbola whose asymptotes are the lines 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0 and which passes through origin is

$12 x^{2} - 7 x y - 12 y^{2} + 31 x + 17 y = 0$

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$12 x^{2} - 7 x y + 12 y^{2} + 31 x + 17 y = 0$

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$12 x^{2} - 7 x y + 12 y^{2} + 31 x + 17 y = 0$

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

$12 x^{2} + 7 x y - 12 y^{2} + 31 x + 17 y = 0$

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Solution

The correct option is A
$12 x^{2} - 7 x y - 12 y^{2} + 31 x + 17 y = 0$

Equation of Hyperbola is (3x – 4y + 7) (4x + 3y + 1) + $λ$ = 0

passes through orgin $\Rightarrow$ 7 + $λ$ = 0 $\Rightarrow$ $λ$ = -7

∴ (3x – 4y + 7) (4x + 3y + 1) – 7 = 0 $\Rightarrow$ $12 x^{2} - 7 x y - 12 y^{2} + 31 x + 17 y = 0$

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Similar questions

Q. The equation of the hyperbola whose asymptotes are the straight lines 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0 and which passes through the origin, is

Q. The equation of the hyperbola whose asymptotes are straight lines

3 x - 4 y + 7 = 0

and

4 x + 3 y + 1 = 0

and which passes through the origin is :

Q. Assertion :The equation of hyperbola whose asymptotes are the lines

3 x - 4 y + 7 = 0

4 x + 3 y + 1 = 0

& which passes through origin is

(3 x - 4 y + 7) (4 x + 3 y + 1) = 7

. Reason: The equation of the pair of asymptotes & hyperbola differ by a constant only.

Q. The equation of the hyperbola whose asymptotes are the straight lines

3 x - 4 y + 7 = 0

and

4 x + 3 y + 1 = 0

and
passing through the origin, is:

A hyperbola passing through origin has 3x – 4y – 1=0 and 4x – 3y – 6 = 0 as its asymptotes. Then the equations of its transverse and conjugate axes are

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The equation of the hyperbola whose asymptotes are the lines 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0 and which passes through origin is

The correct option is A 12x2−7xy−12y2+31x+17y=0 Equation of Hyperbola is (3x – 4y + 7) (4x + 3y + 1) + λ = 0 passes through orgin ⇒ 7 + λ = 0 ⇒ λ = -7 ∴ (3x – 4y + 7) (4x + 3y + 1) – 7 = 0 ⇒ 12x2–7xy–12y2+31x+17y=0