If a normal drawn to the ellipse x^2 4 + y^2 3 =1 touches the hyperbola x^2 4 - y^2 3 =1 , then the square of the slope of that normal is

Step-by-step solution

For normal of ellipse: x^2 4 + y^2 3 =1 ...(i) y^ =- 3 4 x y Slope of normal is m=- 1 y^ = 4 3 y x ...(ii) For tangent of hyperbola: x^2 4 - y^2 3 =1 ...(iii) y^ = 3 4 x y Slope of tangent is m^ =y^ = 3 4 x y ...(iv) Normal of ellipse touches the hyperbola. So it becomes tangent to hyperbola. m=m^ 4 3 y x = 3 4 x y 9 x^2=16 y^2 ...(v) Using above value in equation (iii) 16 y^2 9 4 - y^2 3 =1 y^2=9 From eq ^ n (iii) x^2 4 - 9 3 =1 x^2=16 Now, square of the slope of the normal is m^2= 16 9 y^2 x^2 = 16 9 9 16 =1