If A and B are the points of intersection of the circle x^2 + y^2 - 8x = 0 and the hyperbo…

Question

If A and B are the points of intersection of the circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ and a point P moves on the line $2x - 3y + 4 = 0$, then the centroid of $\Delta PAB$ lies on the line :

Accepted Answer

$$\begin{aligned}
& x^2+y^2-8 x=0, \frac{x^2}{9}-\frac{y^2}{4}=1 \quad\text{.... (1)}\
& 4 x^2-9 y^2=36 \quad\text{.... (2)}\
& \text { Solve }(1) &(2) \
& 4 x^2-9\left(8 x-x^2\right)=36 \
& 13 x^2-72 x-36=0 \
& (13 x+6)(x=6)=0 \
& x=\frac{-6}{13}, x=6 \
& x=\frac{-6}{13}(\text { rejected }) \
& y \rightarrow \text { Imaginary } \
& n=6, \frac{36}{9}-\frac{y^2}{4}=1 \
& y^2=12, y=I \sqrt{12} \
& A(6, \sqrt{12}), B(6,-\sqrt{12}) \
& p\left(\alpha, \frac{2 \alpha+4}{3}\right) P \text { lies on }
\end{aligned}$$

$$\begin{aligned}
& \text { centroid }(\mathrm{h}, \mathrm{k}) \quad 2x-3y+y=0\
& \mathrm{h}=\frac{12+\alpha}{3}, \alpha=3 \mathrm{~h}-12 \
& \mathrm{k}=\frac{\frac{2 \alpha-3 y}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{y} \
& \alpha=\frac{9 \mathrm{k}-4}{2} \
& 6 \mathrm{~h}-2 \mathrm{y}=9 \mathrm{k}-4 \
& 6 \mathrm{x}-9 \mathrm{y}=20
\end{aligned}$$

Solutions · 1