If the ellipse x²/a²+y²/b²=1 meets the line x/7+y/(2 √6)=1 on the x-axis and the line x/7-y/(2 √6)=1 on the y-axis, then the eccentricity of the ellipse is :

Solution

Maths sir · 4 steps

Each line touches the ellipse right where it crosses an axis, so read off a and b from the intercepts.

1. Point on x-axis. Put y = 0 in x/7+y/2√6=1 to get x = 7. The ellipse passes through (7, 0), and it crosses the x-axis only at (± a, 0), so a = 7.
2. Point on y-axis. Put x = 0 in x/7-y/2√6=1 to get y = -2√6. The ellipse crosses the y-axis only at (0, ± b), so b = 2√6 and b² = 24.
3. Find eccentricity. Here a² = 49 > b² = 24, so the major axis is along x and b² = a²(1 - e²).

   e² = 1 - b²/a² = 1 - 24/49 = 25/49

4. So. e = 5/7, which is option (a).

The idea to keep. When a curve meets a line on an axis, put the other coordinate to zero in the line to get the point. Then check which of a², b² is bigger before you write e² = 1 - smaller/bigger.

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