Ellipse

Concept:The acute angle between two tangents at the intersection point of the ellipses is found using the slopes of the tangents and the formula $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$.Explanation:First find the intersection point of $x^2+4y^2=1$ and $4x^2+y^2=1$ in the first quadrant.Subtract the equations: $(x^2+4y^2)-(4x^2+y^2)=0$ gives $3x^2-3y^2=0$, so $x^2=y^2$.Since $P$ lies in the first quadrant, $x=y>0$. Substitute $x=y$ into $x^2+4y^2=1$: $x^2+4x^2=5x^2=1$, so $x=\frac{1}{\sqrt5}$ and $y=\frac{1}{\sqrt5}$.Thus $P\left(\frac{1}{\sqrt5},\frac{1}{\sqrt5}\right)$.Now write the tangent to ellipse $E_1: x^2+4y^2=1$ at $P$. Using the formula $T=0$: $x\cdot\frac{1}{\sqrt5}+4y\cdot\frac{1}{\sqrt5}=1$. Multiplying by $\sqrt5$ gives $x+4y=\sqrt5$. Its slope is $m_1=-\frac{1}{4}$.Similarly, the tangent to $E_2: 4x^2+y^2=1$ at $P$: $4x\cdot\frac{1}{\sqrt5}+y\cdot\frac{1}{\sqrt5}=1$ simplifies to $4x+y=\sqrt5$. Its slope is $m_2=-4$.Compute the acute angle $\theta$ between the tangents: $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| = \left|\frac{-\frac14 - (-4)}{1+(-\frac14)(-4)}\right| = \left|\frac{-\frac14+4}{1+1}\right| = \frac{\frac{15}{4}}{2} = \frac{15}{8}$.Then $4\tan\theta = 4 \times \frac{15}{8} = \frac{15}{2} = 7.5$.Answer:$4\tan\theta = 7.5$ (or $\frac{15}{2}$).