Question

Equation of the hyperbola passing through the point $(1,-1)$ and having asymptotes $\text{x}$ + $2$$\text{y}$+ $3$ = $0$ and $3$$\text{x}$ + $4$$\text{y}$ + $5$ = $0$ is

• A. $3x^2$ + $10$$\text{x}y$ + $8$$y^2$ + $14$$\text{x}$ + $22$$\text{y}$ + $7$ = $0$
• B. $3x^2$ - $10$$\text{x}y$ + $8$$y^2$ + $14$$\text{x}$ + $22$$\text{y}$ + $7$ = $0$
• C. $3x^2$ + $10$$\text{x}y$ + $8$$y^2$ - $14$$\text{x}$ + $22$$\text{y}$ + $7$ = $0$
• D. None of these

Answer 1

Answer: (A)

$3x^2$ + $10$$\text{x}y$ + $8$$y^2$ + $14$$\text{x}$ + $22$$\text{y}$ + $7$ = $0$

Given: Hyperbola passes through point $(1,-1)$

Having asymptotes: $x+2y+3=0$ and $3x+4y+5=0$

Pair of Asymptotes:$(x+2y+3)\cdot (3x+4y+5)$

$:3x^2+8y^2+10xy+22y+15+14x$ -------------(1)

We know that equation of hyperbola = (Pair of asymptotes) $+$ $\lambda$ represents the family of hyperbola.

Now, family of hyperbola having asymptotes $x+2y+3=0$ and $3x+4y+5=0$

$\Rightarrow H:(x+2y+3)\cdot (3x+4y+5)+\lambda$

$:3x^2+8y^2+10xy+22y+14x+15+\lambda =0$ ---------------(2)

Now as given in question, required hyperbola passes through$(1,-1)$

So, putting $(1,-1)$ in Equation (2)

$\Rightarrow 3(1{)}^2+8(-1{)}^2+10\left(1\right)(-1)+22(-1)+14\left(1\right)+15+\lambda =0$

$\Rightarrow 3+8+(-10)-22+14+15+\lambda =0$

$\Rightarrow \lambda =-8$

Putting the value of $\lambda$ in Equation (2),

$\Rightarrow 3x^2+8y^2+10xy+22y+14x+15-8=0$

$\Rightarrow 3x^2+10xy+8y^2+14x+22y+7=0$ is the required hyperbola.