Question

Let ${}^{\prime }{C}^{\prime }$ be a curve which is locus of the point of the intersection of lines $x=2+m$ and $my=4-m$ . A circle $S\equiv (x-2{)}^2+(y+1{)}^2=25$ intersects the curve $C$ at four points $A,B,C$ and $D$. If $O$ is centre of the curve ${}^{\prime }{C}^{\prime }$ then

• A. eccentricity of $C$ is $\sqrt{3}$
• B. eccentricity of $C$ is $\sqrt{2}$
• C. $O{A}^2+O{B}^2+O{C}^2+O{D}^2=120$
• D. $O{A}^2+O{B}^2+O{C}^2+O{D}^2=100$

Answer 1

Answer: (B), (D)

The correct options are
B eccentricity of $C$ is $\sqrt{2}$
D $O{A}^2+O{B}^2+O{C}^2+O{D}^2=100$

Let intersection point of the given lines will be $(h,k)$
$\therefore (h,k)\equiv (2+m,4/m-1)\Rightarrow (h-2)(k+1)=4$
Hence locus will be
$(x-2)(y+1)=4$, which represents a rectangular hyperbola centred at (2,-1)
Now given circle is
$(x-2{)}^2+(y+1{)}^2=25\Rightarrow r=5$
$\because$ circle intersect the hyperbola at $A,B,C,D$
and $O$ is centre of hyperbola as well as circle
$\therefore OA=OB=OC=OD=r$
$\Rightarrow O{A}^2+O{B}^2+O{C}^2+O{D}^2=4r^2=100$
and eccentricity of $C$ is $\sqrt{2}$