Line joining `A(bcosalpha,bsinalpha)` and `B(acosbeta,asinbeta)`, is produced to the point `M(x,y)` such that `AM:MB=b:a` then `xcos((alpha+beta)/2)+ysin((alpha+beta)/2) =`

Line joining A(b cos alpha,b sin alpha) and B(a cos beta,a sin beta), is produced to the point M(x,y) such that AM:MB=b:a then x cos((alpha+beta)/(2))+y sin((alpha+beta)/(2))=

The line joining A(b cos alpha b sin alpha) and B(a cos beta,a sin beta) is produced to the point M(x,y) so that AM and BM are in the ratio b:a. Then prove that x+y tan(alpha+(beta)/(2))=0

Distance between the points A(acosalpha, "asin alpha) and B(acosbeta," asinbeta) is equal to

s(theta-alpha)=a and cos(theta-beta)=b, then sin^(2)(alpha-beta)+2ab cos(alpha-beta)=

If bsinalpha=asin(alpha+2beta) , then (a+b)/(a-b)=?

Under rotation of axes through theta , x cosalpha + ysinalpha=P changes to Xcos beta + Y sin beta=P then . (a) cos beta = cos (alpha - theta) (b) cos alpha= cos( beta - theta) (c) sin beta = sin (alpha - theta) (d) sin alpha = sin ( beta - theta)

If (x) / (a) cos alpha + (y) / (b) sin alpha = 1, (x) / (a) cos beta + (y) / (b) sin beta = 1 and (cos alpha cos beta) / (a ^ (2)) + (without alpha without beta) / (b ^ (2)) = 0 then

If alpha-beta= constant,then the locus of the point of intersection of tangents at P(a cos alpha,b sin alpha) and Q(a cos beta,b sin beta) to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is a circle (b) a straight line an ellipse (d) a parabola