1. **Problem 1:** Given the hyperbola equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ and points $(4\sqrt{2}, 3)$ and $(8, 3\sqrt{3})$ lie on the locus, find $a$ and $b$. Also check if points $(4,0)$, $(0,3)$, and $(4,3)$ lie on the locus. 2. **Step 1:** Substitute $(4\sqrt{2}, 3)$ into the hyperbola equation: $$\frac{(4\sqrt{2})^2}{a^2} - \frac{3^2}{b^2} = 1$$ Simplify: $$\frac{32}{a^2} - \frac{9}{b^2} = 1 \quad (1)$$ 3. **Step 2:** Substitute $(8, 3\sqrt{3})$ into the hyperbola equation: $$\frac{8^2}{a^2} - \frac{(3\sqrt{3})^2}{b^2} = 1$$ Simplify: $$\frac{64}{a^2} - \frac{27}{b^2} = 1 \quad (2)$$ 4. **Step 3:** Let $X = \frac{1}{a^2}$ and $Y = \frac{1}{b^2}$. Rewrite equations (1) and (2): $$32X - 9Y = 1$$ $$64X - 27Y = 1$$ 5. **Step 4:** Solve the system: Multiply the first equation by 3: $$96X - 27Y = 3$$ Subtract the second equation: $$(96X - 27Y) - (64X - 27Y) = 3 - 1$$ $$32X = 2 \Rightarrow X = \frac{1}{16}$$ 6. **Step 5:** Substitute $X=\frac{1}{16}$ into equation (1): $$32 \times \frac{1}{16} - 9Y = 1 \Rightarrow 2 - 9Y = 1 \Rightarrow 9Y = 1 \Rightarrow Y = \frac{1}{9}$$ 7. **Step 6:** Recall $X = \frac{1}{a^2} = \frac{1}{16} \Rightarrow a^2 = 16$ and $Y = \frac{1}{b^2} = \frac{1}{9} \Rightarrow b^2 = 9$. 8. **Step 7:** Check points $(4,0)$, $(0,3)$, and $(4,3)$: - For $(4,0)$: $$\frac{4^2}{16} - \frac{0^2}{9} = \frac{16}{16} - 0 = 1$$ Point lies on the locus. - For $(0,3)$: $$\frac{0^2}{16} - \frac{3^2}{9} = 0 - 1 = -1 \neq 1$$ Point does not lie on the locus. - For $(4,3)$: $$\frac{4^2}{16} - \frac{3^2}{9} = 1 - 1 = 0 \neq 1$$ Point does not lie on the locus. --- 9. **Problem 2:** Find the equation of the locus of point $P$ such that the ratio of distances to fixed points $A(1,-1)$ and $B(-1,1)$ is $PA : PB = 1 : 2$. 10. **Step 1:** Let $P = (x,y)$. The distances are: $$PA = \sqrt{(x-1)^2 + (y+1)^2}$$ $$PB = \sqrt{(x+1)^2 + (y-1)^2}$$ 11. **Step 2:** Given $\frac{PA}{PB} = \frac{1}{2}$, square both sides: $$\frac{PA^2}{PB^2} = \frac{1}{4}$$ 12. **Step 3:** Substitute distances squared: $$\frac{(x-1)^2 + (y+1)^2}{(x+1)^2 + (y-1)^2} = \frac{1}{4}$$ 13. **Step 4:** Cross-multiply: $$4[(x-1)^2 + (y+1)^2] = (x+1)^2 + (y-1)^2$$ 14. **Step 5:** Expand: $$4(x^2 - 2x + 1 + y^2 + 2y + 1) = x^2 + 2x + 1 + y^2 - 2y + 1$$ $$4x^2 - 8x + 4 + 4y^2 + 8y + 4 = x^2 + 2x + 1 + y^2 - 2y + 1$$ 15. **Step 6:** Simplify: $$4x^2 - 8x + 4y^2 + 8y + 8 = x^2 + 2x + y^2 - 2y + 2$$ 16. **Step 7:** Bring all terms to one side: $$4x^2 - x^2 - 8x - 2x + 4y^2 - y^2 + 8y + 2y + 8 - 2 = 0$$ $$3x^2 - 10x + 3y^2 + 10y + 6 = 0$$ 17. **Step 8:** Complete the square for $x$ and $y$: $$3x^2 - 10x + 3y^2 + 10y = -6$$ Divide by 3: $$x^2 - \frac{10}{3}x + y^2 + \frac{10}{3}y = -2$$ 18. **Step 9:** Complete the square: For $x$: $$x^2 - \frac{10}{3}x = \left(x - \frac{5}{3}\right)^2 - \left(\frac{5}{3}\right)^2 = \left(x - \frac{5}{3}\right)^2 - \frac{25}{9}$$ For $y$: $$y^2 + \frac{10}{3}y = \left(y + \frac{5}{3}\right)^2 - \left(\frac{5}{3}\right)^2 = \left(y + \frac{5}{3}\right)^2 - \frac{25}{9}$$ 19. **Step 10:** Substitute back: $$\left(x - \frac{5}{3}\right)^2 - \frac{25}{9} + \left(y + \frac{5}{3}\right)^2 - \frac{25}{9} = -2$$ $$\left(x - \frac{5}{3}\right)^2 + \left(y + \frac{5}{3}\right)^2 = -2 + \frac{50}{9} = \frac{32}{9}$$ 20. **Final equation:** $$\boxed{\left(x - \frac{5}{3}\right)^2 + \left(y + \frac{5}{3}\right)^2 = \frac{32}{9}}$$ This is a circle with center $\left(\frac{5}{3}, -\frac{5}{3}\right)$ and radius $\frac{4\sqrt{2}}{3}$. --- **Summary:** - $a^2 = 16$, $b^2 = 9$ for the hyperbola. - Points $(4,0)$ lies on the hyperbola; $(0,3)$ and $(4,3)$ do not. - The locus with distance ratio 1:2 is a circle centered at $\left(\frac{5}{3}, -\frac{5}{3}\right)$ with radius $\frac{4\sqrt{2}}{3}$.