Coordinate Geometry
Hyperbola
MMTS_Full_Test_12
Grade 12
Question:
A variable chord of the hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{8}=1$ subtends a right angle at the centre. If this chord touches a fixed circle which is concentric with the hyperbola and $r$ is radius of the circle, then $r^2$ is
Step-by-Step Solution
Key Concept: Chord $lx+my=1$ with right-angle condition; find envelope
Chord $lx+my=1$ subtends right angle at origin: combined equation gives $\frac{x^2}{4}-\frac{y^2}{8}-(lx+my)^2=0$. Coeff of $x^2$ + coeff of $y^2=0\Rightarrow\frac{1}{4}-l^2-\frac{1}{8}-m^2=0\Rightarrow l^2+m^2=1/8$. Dist from origin to chord $=\frac{1}{\sqrt{l^2+m^2}}=\sqrt{8}$. $r^2=8$.
Correct Answer: 8