Question 17

Step 1: Find the intersection point $P$ in the first quadrant.

The equations are $x^2 + 4y^2 = 1$ and $4x^2 + y^2 = 1$.

Subtracting the two equations:

$(4x^2 + y^2) - (x^2 + 4y^2) = 1 - 1$

$3x^2 - 3y^2 = 0 \implies x^2 = y^2$.

Since $P$ is in the first quadrant, $x = y$.

Substituting $x = y$ into $x^2 + 4y^2 = 1$:

$x^2 + 4x^2 = 1 \implies 5x^2 = 1 \implies x = \frac{1}{\sqrt{5}}$ and $y = \frac{1}{\sqrt{5}}$. So, $P = (\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}})$.

Step 2: Find the slopes of the tangents at $P$.

Differentiating $x^2 + 4y^2 = 1$: $2x + 8y y' = 0 \implies m_1 = -\frac{x}{4y}$. At $P$, $m_1 = -\frac{1/\sqrt{5}}{4/\sqrt{5}} = -\frac{1}{4}$.

Differentiating $4x^2 + y^2 = 1$: $8x + 2y y' = 0 \implies m_2 = -\frac{4x}{y}$. At $P$, $m_2 = -\frac{4/\sqrt{5}}{1/\sqrt{5}} = -4$.

Step 3: Calculate $\tan \theta$.

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-1/4 - (-4)}{1 + (-1/4)(-4)} \right| = \left| \frac{15/4}{1 + 1} \right| = \frac{15}{8}$.

Step 4: Find $4 \tan \theta$.

$4 \tan \theta = 4 \times \frac{15}{8} = \frac{15}{2} = 7.5$.