(y-b)^(2)=4(x-a)

If(x+a)^(2)+(y+b)^(2)=4(ax+by), where x,a,y,b are real,the value of xy-ab is :

(y -b)^(2) = 4k (x - a)

Find the area enclosed between the parabolas y^(2)=4b(b-x) and y^(2)=4a(x+a) .

The differential equation whose solution is (y - b)^(2) = 4 k (x - a) (where b,a,k are constants ) is of

For any non-zero real value of "m" ,the equation of the parabola to which the line " mx-y+10+m^(2)=0 " is a tangent,is (A) " x^(2)=y-10 (B) " y^(2)=4(x-2) (C) " x^(2)=-4(y-10) (D) x^(2)=-4y

Show that the curves y^(2)=4a(x+a) and y^(2)=4b(b-x)(a gt ,b gt 0) intresect orthogonally.

Show that the area included between the parabolas y^(2)=4a(x+a) and y^(2)=4b(b-x) is (8)/(3)sqrt(ab)(a+b)