Question

Area of the quadrilateral with its vertices at foci of the conics $9x^2-16y^2-18x+32y-23=0$ and $25x^2+9y^2-50x-18y+33=0$ is

• A. $\frac{5}{6}$
• B. $\frac{8}{9}$
• C. $\frac{5}{3}$
• D. $\frac{16}{9}$

Answer 1

The correct option is B

$\frac{8}{9}$

Conic I : $9x^2-16y^2-18x+32y-23=0$

$=9x^2-18x-(16y^2-32y)-23=0$

$=9(x^2-2x)-16(y^2-2y)-23=0$

$=9(x^2-2x+1-1)-16(y^2-2y+1-1)-23=0$

$=9(x-1{)}^2-9-16(y-1{)}^2+16-23=0$

$=9(x-1{)}^2-16(y-1{)}^2=16$

$=\frac{(x-1{)}^2}{\frac{16}{9}}-\frac{(y-1{)}^2}{1}=1$ So, conic I is Hyperbola.

Comparing it with the standard equation of Hyperbola: $\frac{X^2}{a^2}-\frac{Y^2}{b^2}=1$

We get, $a^2=\frac{16}{9}\Rightarrow a=\pm \frac{4}{3}$ and $b^2=1\Rightarrow b=\pm 1$

Since, vertex of a standard hyperbola is $X=0,Y=0$

hence, vertex of our hyperbola is $x-1=0,y-1=0\Rightarrow x=1,y=1$ i.e., $(1,1)$

conic II : $25x^2+9y^2-50x-18y+33=0$

$=25x^2-50x+9y^2-18y+33=0$

$=25(x^2-2x+1-1)+9(y^2-2y+1-1)+33=0$

$=25(x-1{)}^2-25+9(y-1{)}^2-9+33=0$

$=25(x-1{)}^2+9(y-1{)}^2=1$

$=\frac{(x-1{)}^2}{\frac{1}{25}}+\frac{(y-1{)}^2}{\frac{1}{9}}=1$ So, conic II is Ellipse.

Comparing it with the standard equation of Ellipse: $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$

We get, $a^2=\frac{1}{25}\Rightarrow a=\pm \frac{1}{5}$ and $b^2=\frac{1}{9}\Rightarrow b=\pm \frac{1}{3}$

Since, vertex of a standard ellipse is $X=0,Y=0$

hence, vertex of our ellipse is $x-1=0,y-1=0\Rightarrow x=1,y=1$ i.e., $(1,1)$

So, we have to find out the area of the quadrilateral formed by joining the foci $f_1,f_2,f_3$ and $f_4$

Foci of conic I are $(\pm \sqrt{a^2+b^2},0)$

$=(\pm \sqrt{\frac{16}{9}+1,0})$

$=(\pm \frac{5}{3},0)$

$\Rightarrow f_1=(\frac{5}{3},0)$ and $f_2=(-\frac{5}{3},0)$

The ellipse has its major axis along y-axis, so, its foci are $(0,\pm \sqrt{b^2-a^2})$

$=(0,\pm \sqrt{\frac{1}{9}-\frac{1}{25}})$

$=(0,\pm \sqrt{\frac{16}{9\times 25}})$

$=(0,\pm \frac{4}{15})$

$\Rightarrow f_3=(0,\frac{4}{15})$ and $f_4=(0,-\frac{4}{15})$

Clearly, $f_1{f}_3{f}_2{f}_4$ is a rhombus and its area won't change even if the vertex is changed from $(0,0)$ to $(1,1)$

We know that area of a rhombus is $\frac{1}{2}\times \text{Product of diagonals}$

Let the diagonals be $d_1$ and $d_2$, where, $d_1=f_1{f}_2=\frac{5}{3}+\frac{5}{3}=\frac{10}{3}$

and $d_2=f_3{f}_4=\frac{4}{15}+\frac{4}{15}=\frac{8}{15}$

$\Rightarrow$ Area $=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times \frac{10}{3}\times \frac{8}{15}=\frac{8}{9}$