JEE Main202621 January 2026Evening ShiftMathematicsEllipseActual
If the line x+4 y= 7 , where R , touches the ellipse 3 x²+4 y²=1 at the point P in the first quadrant, then one of the focal distances of P is :
Options
- A1 3 + 1 2 7
- B1 3 - 1 2 5
- C1 3 + 1 2 5
- D1 3 - 1 2 11
Correct answer
A. 1 3 + 1 2 7
Step-by-step solution
Ellipse 3x^2 + 4y^2 = 1 : a^2 = 1 3 , b^2 = 1 4 , e = 1 2 . Tangent at (x₀, y₀) : 3x₀x + 4y₀y = 1 . Comparing with x + 4y = 7 : 4y₀ = 4 7 y₀ = 1 7 and 3x₀ = 7 . Substituting in ellipse equation: ^2 21 + 4 7 = 1 ^2 = 9 = 3 . P = ( 1 7 , 1 7 ) . Focal distances = a ex₀ = 1 3 1 2 7 . One focal distance is 1 3 + 1 2 7 .