Which one of the following curves cut the parabola at right angles? `x^2+y^2=a^2` (b) `y=e^(-x//2a)` `y=a x` (d) `x^2=4a y`

Which of the following pair(s) of curves is/are orthogonal? y^2=4a x ; y=e^(-x/(2a)) y^2=4a x ; x^2=4a ya t(0,0) x y=a^2; x^2-y^2=b^2 y=a x ;x^2+y^2=c^2

Which of the following points lie on the parabola x^2=4a y ? a. x=a t^2, y=2a t b. x=2a t ,\ y=a t^2 c. x=2a t^2, y=a t d. x=2a t ,\ y=a t^2

Which of the following pairs(s) of curves is/are orthogonal? (a) y^2 = 4 a x ; y = e^ (-x/ (2 a)) (b) y^2 = 4 a x ; x^2 = 4ay (c) xy = a^2 ; x^2-y^2 = b^2 (d) y = a x ; x^2 + y^2 = c^2

Find the angle of intersection of the following curves : y^2=x \ a n d \ x^2=y

Find the condition for the line y=m x to cut at right angles the conic a x^2+2h x y+b y^2=1.

The axis of a parabola is along the line y=x and the distance of its vertex and focus from the origin are sqrt(2) and 2sqrt(2) , respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is (a) (x+y)^2=(x-y-2) (b) (x-y)^2=(x+y-2) (c) (x-y)^2=4(x+y-2) (d) (x-y)^2=8(x+y-2)

Which of the following differential equations has y = x as one of its particular solution? (A) (d^2y)/(dx^2)-x^2(dy)/(dx)+x y=x (B) (d^2y)/(dx^2)+x(dy)/(dx)+x y=x (C) (d^2y)/(dx^2)-x^2(dy)/(dx)+x y=0 (D) (d^2y)/(dx^2)+x(dy)/(dx)+x y=0

Find the equation of line which is normal to the parabola x^(2)=4y and touches the parabola y^(2)=12x .

The equation of common tangent(s) to the parabola y^2=16 x and the circle x^2+y^2=8 is/are x+y+4=0 (b) x+y-4=0 x-y-4=0 (d) x-y+4=0

Show that the curves 4x=y^2 and 4x y=k cut at right angles, if k^2=512 .