For the hyperbola H: x²-y²=1 and the ellipse E: x²/a²+y²/b²=1, a >b>0, let the
(1) eccentricity of E be reciprocal of the eccentricity of H, and
(2) the line y=√(5/2) x+K be a common tangent of E and H.
Then 4(a²+b²) is equal to _____________.

Solution

Maths sir · 4 steps

Use each curve's tangency condition c² = … for the same line, then equate the two values of K².

1. Link the eccentricities. For x² - y² = 1, e_H = √(1 + 1) = √2, so e_E = 1/√2. Then b² = a²(1 - e_E²) = a²/2.
2. Tangent to hyperbola. The line y = mx + c touches x²/A² - y²/B² = 1 when c² = A²m² - B². With A² = B² = 1 and m² = 5/2, this gives K² = 5/2 - 1 = 3/2.

   K² = a² m² + b² = 5/2a² + a²/2 = 3a²

3. Tangent to ellipse. For the ellipse the condition is c² = a²m² + b², which gives K² = 3a². The same K touches both curves, so 3a² = 3/2. That gives a² = 1/2 and b² = 1/4.
4. So. 4(a² + b²) = 4(1/2 + 1/4) = 4 × 3/4 = 3.

The common slip: 7. Using the ellipse condition c² = a²m² + b² for the hyperbola gives K² = 7/2. That leads to a² = 7/6 and the answer 7. The hyperbola needs a minus sign.

The idea to keep. For a common tangent of given slope, write c² from each curve's tangency condition and equate them. The ellipse uses a²m² + b², the hyperbola uses a²m² - b².

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