1. Problem 14: Find values of $a,b,c,d$ such that the line $y=x+1$ is a common tangent to the curves $$x^2 + y^2 + ax + by = 1$$ and $$x^2 + y^2 + cx + dy = -8$$ at the point $(1,2)$. 2. Since $(1,2)$ lies on both curves and the line, substitute $x=1$, $y=2$: $$1^2 + 2^2 + a(1) + b(2) = 1 \implies 1 + 4 + a + 2b = 1 \implies a + 2b = 1 - 5 = -4$$ $$1^2 + 2^2 + c(1) + d(2) = -8 \implies 1 + 4 + c + 2d = -8 \implies c + 2d = -8 - 5 = -13$$ 3. The line $y = x + 1$ has slope $m=1$. The tangent to each curve at $(1,2)$ must have slope 1. 4. Differentiate implicitly each curve: For $x^2 + y^2 + ax + by = 1$: $$2x + 2y \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$$ Rearranged: $$\frac{dy}{dx}(2y + b) = -2x - a \implies \frac{dy}{dx} = \frac{-2x - a}{2y + b}$$ At $(1,2)$, slope = 1: $$1 = \frac{-2(1) - a}{2(2) + b} = \frac{-2 - a}{4 + b}$$ Cross multiply: $$4 + b = -2 - a$$ Rearranged: $$a + b = -6$$ 5. For the second curve $x^2 + y^2 + cx + dy = -8$: Similarly, $$2x + 2y \frac{dy}{dx} + c + d \frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = \frac{-2x - c}{2y + d}$$ At $(1,2)$, slope = 1: $$1 = \frac{-2(1) - c}{2(2) + d} = \frac{-2 - c}{4 + d}$$ Cross multiply: $$4 + d = -2 - c$$ Rearranged: $$c + d = -6$$ 6. Now solve the system: $$\begin{cases} a + 2b = -4 \\ a + b = -6 \end{cases}$$ Subtract second from first: $$a + 2b - (a + b) = -4 - (-6) \implies b = 2$$ Then from $a + b = -6$: $$a + 2 = -6 \implies a = -8$$ 7. Similarly for $c,d$: $$\begin{cases} c + 2d = -13 \\ c + d = -6 \end{cases}$$ Subtract second from first: $$c + 2d - (c + d) = -13 - (-6) \implies d = -7$$ Then from $c + d = -6$: $$c - 7 = -6 \implies c = 1$$ 8. Final answer: $$a = -8, b = 2, c = 1, d = -7$$ --- 9. Problem 15: Find tangent lines to curve $$y = \frac{x}{x+1}$$ parallel to line $y = x$. 10. The slope of $y = x$ is 1. Find $\frac{dy}{dx}$ and set equal to 1. 11. Differentiate: $$y = \frac{x}{x+1} = x(x+1)^{-1}$$ Using quotient or product rule: $$\frac{dy}{dx} = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{x + 1 - x}{(x+1)^2} = \frac{1}{(x+1)^2}$$ 12. Set slope equal to 1: $$\frac{1}{(x+1)^2} = 1 \implies (x+1)^2 = 1 \implies x+1 = \pm 1$$ 13. Solve for $x$: $$x+1=1 \implies x=0$$ $$x+1=-1 \implies x=-2$$ 14. Find corresponding $y$ values: $$y = \frac{x}{x+1}$$ At $x=0$: $$y=0$$ At $x=-2$: $$y = \frac{-2}{-2+1} = \frac{-2}{-1} = 2$$ 15. Equations of tangent lines with slope 1 passing through these points: At $(0,0)$: $$y - 0 = 1(x - 0) \implies y = x$$ At $(-2,2)$: $$y - 2 = 1(x + 2) \implies y = x + 4$$ 16. For the second part, given $$x = a \cos \phi, \quad y = b \sin \phi$$ Show that $$\frac{dy}{dx} = - \cot \phi$$ 17. Differentiate parametric equations: $$\frac{dy}{d\phi} = b \cos \phi$$ $$\frac{dx}{d\phi} = -a \sin \phi$$ 18. Then $$\frac{dy}{dx} = \frac{\frac{dy}{d\phi}}{\frac{dx}{d\phi}} = \frac{b \cos \phi}{-a \sin \phi} = - \frac{b}{a} \cot \phi$$ 19. If $a = b$, then $$\frac{dy}{dx} = - \cot \phi$$ --- 20. Problem 13: Curves $$x^2 + y^2 = 10$$ and $$x^2 + y^2 + a y = b$$ intersect at right angles at $x=3$. Find $a,b$. 21. At $x=3$, find $y$ on first curve: $$3^2 + y^2 = 10 \implies y^2 = 1 \implies y = \pm 1$$ 22. The curves intersect at $(3,1)$ or $(3,-1)$. 23. Differentiate both curves implicitly: First curve: $$2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$$ Second curve: $$2x + 2y \frac{dy}{dx} + a \frac{dy}{dx} = 0 \implies (2y + a) \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = \frac{-2x}{2y + a}$$ 24. At intersection, slopes are negative reciprocals (right angle): $$m_1 \cdot m_2 = -1$$ 25. Substitute slopes: $$\left(-\frac{x}{y}\right) \cdot \frac{-2x}{2y + a} = -1$$ Simplify: $$\frac{2x^2}{y(2y + a)} = -1$$ 26. At $x=3$, $y=1$: $$\frac{2 \cdot 9}{1(2 \cdot 1 + a)} = -1 \implies \frac{18}{2 + a} = -1$$ Cross multiply: $$18 = - (2 + a) \implies 18 = -2 - a \implies a = -20$$ 27. Find $b$ by substituting $(3,1)$ into second curve: $$3^2 + 1^2 + (-20)(1) = b \implies 9 + 1 - 20 = b \implies b = -10$$ 28. Check for $y = -1$: $$\frac{2 \cdot 9}{-1(2(-1) + a)} = -1 \implies \frac{18}{-1(-2 + a)} = -1 \implies \frac{18}{2 - a} = -1$$ $$18 = - (2 - a) \implies 18 = -2 + a \implies a = 20$$ 29. Substitute into second curve: $$9 + 1 + 20(-1) = b \implies 10 - 20 = b \implies b = -10$$ 30. Final answers: $$a = -20, b = -10$$ or $$a = 20, b = -10$$