Question

The intersection of the hyperbola $\frac{(x+1{)}^2}{8}-\frac{(y-1{)}^2}{9}=1$ and the ellipse $\frac{(x+1{)}^2}{32}+\frac{(y-1{)}^2}{18}=1$ is

• A. $(3,4),(-3,-4),(5,4),(-5,-4)$
• B. $(3,2),(-3,-2),(5,2),(-5,-2)$
• C. $(3,4),(3,-2),(-5,4),(-5,-2)$
• D. $(3,4),(3,2),(5,4),(5,2)$

Answer 1

The correct option is C $(3,4),(3,-2),(-5,4),(-5,-2)$

Multiply both sides of the equation $\frac{(x+1{)}^2}{32}+\frac{(y-1{)}^2}{18}=1$ by $2$ to get $\frac{(x+1{)}^2}{16}+\frac{(y-1{)}^2}{9}=2$.

Add this to $\frac{(x+1{)}^2}{8}-\frac{(y-1{)}^2}{9}=1$ by $2$ to get $\frac{3(x+1{)}^2}{16}=3$, so that

$(x+1{)}^2=16$

$\Rightarrow x+1=\pm 4\Rightarrow x+1=4,x+1=-4\Rightarrow x=3,x=-5$

Substitute $x=3$ in $\frac{(x+1{)}^2}{16}+\frac{(y-1{)}^2}{9}=2$ that is:

$\frac{(3+1{)}^2}{16}+\frac{(y-1{)}^2}{9}=2$

$\Rightarrow \frac{4^2}{16}+\frac{(y-1{)}^2}{9}=2$

$\Rightarrow \frac{16}{16}+\frac{(y-1{)}^2}{9}=2$

$\Rightarrow 1+\frac{(y-1{)}^2}{9}=2$

$\Rightarrow \frac{(y-1{)}^2}{9}=2-1$

$\Rightarrow \frac{(y-1{)}^2}{9}=1$

$\Rightarrow (y-1{)}^2=9$

$\Rightarrow y-1=\pm 3\Rightarrow y-1=3,y-1=-3\Rightarrow y=4,y=-2$

Hence, the coordinates of the four points of intersection for these two graphs are:

$(3,4),(3,-2),(-5,4),(-5,-2)$