A common tangent T to the curves C₁: x²/4+y²/9=1 and C₂: x²/42-y²/143=1 does not pass through the fourth quadrant. If T touches C₁ at (x₁, y₁) and C₂ at (x₂, y₂), then |2 x₁+x₂| is equal to ______________.

Solution

Maths sir · 4 steps

Write the tangent as y = mx + c, use both tangency conditions, and let the quadrant rule pick the sign.

1. Tangency conditions. The line y = mx + c touches x²/a² + y²/b² = 1 when c² = a²m² + b². It touches x²/a² - y²/b² = 1 when c² = a²m² - b². For a common tangent, both values of c² must be equal.

   4m² + 9 = 42m² - 143 ⇒ 38m² = 152 ⇒ m² = 4, c² = 25

2. Pick the line. With m = -2, y becomes negative for large positive x, so the line enters the fourth quadrant. With m = 2 and c = -5, the point (1, -3) lies on it, so that line also enters the fourth quadrant. Only y = 2x + 5 stays out, because y > 5 whenever x > 0.
3. Points of contact. Rewrite T as -2x/5 + y/5 = 1. Match it with the ellipse tangent xx₁/4 + yy₁/9 = 1 to get x₁ = -8/5. Match it with the hyperbola tangent xx₂/42 - yy₂/143 = 1 to get x₂ = -84/5.
4. So. 2x₁ + x₂ = -16/5 - 84/5 = -20, so |2x₁ + x₂| = 20.

The idea to keep. For a common tangent to two conics, equate the two c² conditions to get the slope. Then compare T with the T = 1 tangent form of each curve to read off the contact points.

Answer from the NTA final answer key. Solution by PrepWiser.