Question

Let the normal at (2cosα,sinα) to the ellipse x^2+4y^2=4 intersects the ellipse again at (2cosβ,sinβ)(α,β∈R). If sinβ=−sinα(a+bsin2α)  / ( c+dsin2α) in simplest form, then the value of (a+b-c-d) is equal to (a,b,c,d∈I)

Answer 1

0 is the answer 

Answer 2

The equation of normal to the ellipse x² + 4y² = 4 at (2cosα, sinα) is 

\(\frac{4x}{2cos\alpha} - \frac y{sin \alpha} = 4 -1\)

⇒ \(2xsec\alpha - ycosec\alpha = 3\)

It passes through \((2cos\beta, sin \beta).\)

\(\therefore 4cos\beta \,sec \alpha - sin\beta\, cosec \alpha = 3\)

⇒ \(4 cos \beta \, sin \alpha - sin \beta \, cos \alpha = 3 sin \alpha \, cos \alpha\)

⇒ \(sin \beta \, cos \alpha = 4 cos \beta \, sin \alpha - 3sin\alpha \, cos \beta \)

\(= sin \alpha (4 cos \beta -3cos \alpha)\)

\(= - sin \alpha (3cos \alpha - 4cos \beta)\)

\(\therefore sin\beta = -sin\alpha\left(\frac{3cos \alpha - 4cos \beta}{cos\alpha}\right)\)

\(= -sin\alpha\left(\frac{3cos^2 \alpha - 4cos \alpha\,cos\beta}{cos^2\alpha}\right)\)

\(= -sin\alpha\left(\frac{3-3sin^2 \alpha - 4cos \alpha\,cos\beta}{1-sin^2\alpha}\right)\)

\(\therefore a =3, b = -3, c= 1 , d=-1\)

\(\therefore a + b - c - d = 3 + (-3) - 1 - (-1) =0\)