Answer
See explanations.
Work Step by Step
The pairs of curves will meet orthogonally if the tangents to the curves at their intersection point are normal to each other.
(a) Step 1. Find the intersection point: $3y^2+y^2=4$, $y^2=1$, $y=\pm1$, $x^2=3$, $x=\pm\sqrt 3$, which gives points $(\sqrt 3,1)$, $(\sqrt 3,-1)$, $(-\sqrt 3,1)$, $(-\sqrt 3,-1)$,
Step 2. Given the equation $x^2+y^2=4$, differentiate both sides with respect to $x$: $2x+2yy'=0$, which gives $y'=-\frac{x}{y}$; thus the slopes of the tangent line at the above points are $m_i=-\sqrt 3, -\sqrt 3, \sqrt 3, \sqrt 3$
Step 3. Given the equation $x^2=3y^2$, differentiate both sides with respect to $x$: $2x=6yy'$, which gives $y'=\frac{x}{3y}$; thus the slopes of the tangent lines at the above points are $n_i=\sqrt 3/3, \sqrt 3/3, -\sqrt 3/3, -\sqrt 3/3$
Step 4. Check normality: $m_in_i=-1$
(b) Step 1. Find the intersection point: $y^2/3=1-y^2$, $y^2=3/4$, $y=\pm\frac{\sqrt 3}{2}$, $x=\pm\frac{1}{4}$, which gives points $(\frac{1}{4},\frac{\sqrt 3}{2}$ and $(\frac{1}{4},-\frac{\sqrt 3}{2}$
Step 2. Given the equation $x=1-y^2$, differentiate both sides with respect to $x$: $1=-2yy'$, which gives $y'=-\frac{1}{2y}$, thus the slopes of the tangent lines at the above points are $m_1=-\frac{\sqrt 3}{3}$ and $m_2=\frac{\sqrt 3}{3}$
Step 3. Given the equation $x=y^2/3$, differentiate both sides with respect to $x$: $1=2yy'/3$, which gives $y'=\frac{3}{2y}$; thus the slopes of the tangent lines at the above points are $n_1=\sqrt 3$ and $n_2=-\sqrt 3$
Step 4. Check normality: $m_1n_1=-1$ and $m_2n_2=-1$
