Let a line L pass through the point of intersection of the lines b x+10 y-8=0 and 2 x-3 y=0, b ∈ R-{4/3}. If the line L also passes through the point (1,1) and touches the circle 17(x²+y²)=16, then the eccentricity of the ellipse x²/5+y²/b²=1 is :

Solution

Maths sir · 5 steps

Use the family of lines through the intersection point, then let the tangency condition fix b².

1. Family of lines. Any line through the intersection of the two given lines can be written as bx + 10y - 8 + λ(2x - 3y) = 0. It passes through (1, 1), so b + 2 - λ = 0, which gives λ = b + 2.
2. Equation of L. Substitute λ = b + 2 and collect terms: (3b + 4)x + (4 - 3b)y - 8 = 0.
3. Tangency condition. The circle is x² + y² = 16/17, with centre at the origin and radius 4/√17. L touches the circle, so the distance from the origin to L equals this radius.

   8/√((3b+4)² + (4-3b)²) = 4/√17 ⇒ 18b² + 32 = 68 ⇒ b² = 2

4. Eccentricity. The ellipse is x²/5 + y²/2 = 1. Here 5 > 2, so the major axis lies along the x-axis, with a² = 5 and b² = 2. Then e² = 1 - 2/5 = 3/5.
5. So. e = √(3/5), which is option (b).

The idea to keep. A line through the meeting point of two lines is L₁ + λ L₂ = 0. A line touches a circle when its distance from the centre equals the radius.

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