Q.(a) Find the angle between the curves $y=x^2$ and $y=(x-3)^2$. OR

(a) Finds the intersection of the two parabolas, computes each curve's slope there, and applies the angle-between-lines formula; (b) combines the two arctangents via the addition formula and solves the resulting equation in $x$. Both alternatives answered below.

(a) Angle between $y=x^2$ and $y=(x-3)^2$

1. Find the intersection. $x^2=(x-3)^2\Rightarrow x^2=x^2-6x+9\Rightarrow6x=9\Rightarrow x=\dfrac32$. Then $y=\left(\dfrac32\right)^2=\dfrac94$.

2. Slope of each curve at $x=\dfrac32$.

• $y=x^2\Rightarrow y'=2x$; at $x=\dfrac32$: $m_1=3$.
• $y=(x-3)^2\Rightarrow y'=2(x-3)$; at $x=\dfrac32$: $m_2=2\left(\dfrac32-3\right)=2\left(-\dfrac32\right)=-3$.

3. Angle between the curves.

$\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{3-(-3)}{1+3(-3)}\right|=\left|\dfrac{6}{1-9}\right|=\left|\dfrac6{-8}\right|=\dfrac34$

4. $\theta=\tan^{-1}\dfrac34$.

(b) Solve $\tan^{-1}\dfrac{x-1}{x-2}+\tan^{-1}\dfrac{x+1}{x+2}=\dfrac\pi4$

1. Let $A=\dfrac{x-1}{x-2}$, $B=\dfrac{x+1}{x+2}$. Use $\tan^{-1}A+\tan^{-1}B=\tan^{-1}\dfrac{A+B}{1-AB}$ (valid when $AB<1$).

2. Compute $A+B$.

$A+B=\dfrac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}=\dfrac{(x^2+x-2)+(x^2-x-2)}{x^2-4}=\dfrac{2x^2-4}{x^2-4}$

3. Compute $AB$.

$AB=\dfrac{(x-1)(x+1)}{(x-2)(x+2)}=\dfrac{x^2-1}{x^2-4}$

4. Compute $1-AB$.

$1-AB=\dfrac{(x^2-4)-(x^2-1)}{x^2-4}=\dfrac{-3}{x^2-4}$

5. Form $\dfrac{A+B}{1-AB}$. …