Question

Consider the ellipses given by $x^2 + 4y^2 = 1$ and $4x^2 + y^2 = 1$.
Let $P$ be the point in the first quadrant where the given ellipses intersect. If $\theta$ is the acute angle between the tangents to the given ellipses at the point $P$, then the value of $4 \tan \theta$ is {

Hint:

For curves of the form $f(x,y)=C$ and $g(x,y)=C$, if the curves are symmetric reflections across $y=x$, they will always intersect on the line $y=x$ and their slopes will be reciprocals of each other ($m_1 = 1/m_2$).

Answer 1

Correct Answer: 7.5

Step 1: Understanding the Question:
We need to find the intersection point of two symmetric ellipses, find the slopes of the tangents at that point, and then calculate the angle between them.

Step 2: Key Formula or Approach:

• Intersection: solve both equations simultaneously.

• Slope of tangent $m = - \frac{f_x}{f_y}$.

• Angle between lines: $\tan \theta = | \frac{m_1 - m_2}{1 + m_1 m_2} |$.

Step 3: Detailed Explanation:

• Intersection: $x^2 + 4y^2 = 4x^2 + y^2 \implies 3x^2 = 3y^2$. In 1st quadrant, $x = y$.
Substituting into first eq: $x^2 + 4x^2 = 1 \implies 5x^2 = 1 \implies x = \frac{1}{\sqrt{5}}, y = \frac{1}{\sqrt{5}}$.
$P = (1/\sqrt{5}, 1/\sqrt{5})$.

• Tangent to $x^2 + 4y^2 = 1$: Differentiating, $2x + 8y y' = 0 \implies m_1 = -\frac{x}{4y} = -\frac{1}{4}$.

• Tangent to $4x^2 + y^2 = 1$: Differentiating, $8x + 2y y' = 0 \implies m_2 = -\frac{4x}{y} = -4$.

• $\tan \theta = | \frac{-4 - (-1/4)}{1 + (-4)(-1/4)} | = | \frac{-15/4}{1 + 1} | = \frac{15}{8}$.

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

Step 4: Final Answer:
The value of $4 \tan \theta$ is 7.5.