Question

Let $S\equiv 7y^2-9x^2+54x-28y-116=0$ represents a hyperbola. A line $L=0$ is passing through the centre of $S=0$ whose slope is the eccentricity of $S=0$. The sum of the intercepts made by the line $L=0$ on the co-ordinate axes is equal to

• A. $-\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A $-\frac{1}{2}$

Given, $S\equiv 7y^2-9x^2+54x-28y-116=0$
$\Rightarrow 7(y^2-4y+4)-9(x^2-6x+9)=116+28-81$
$\Rightarrow 7(y-2{)}^2-9(x-3{)}^2=63$
$\Rightarrow \frac{(y-2{)}^2}{9}-\frac{(x-3{)}^2}{7}=1$

Now, Centre of $S=0$ is $(3,2)$
Eccentricity of $S=0$ is $\sqrt{1+\frac{7}{9}}=\frac{4}{3}$
Equation of $L:y-2=\frac{4}{3}(x-3)$
$\Rightarrow 3y-6=4x-12$
$\Rightarrow 4x-3y=6$
$\Rightarrow \frac{x}{\left(3/2\right)}+\frac{y}{(-2)}=1$

$\therefore$ Sum of the intercepts $=\frac{3}{2}-2=-\frac{1}{2}$