find the eqn of parabola whose vertex (2,-3) and foci is (0,5)
Answers
Answered by
9
EXPLANATION.
Equation of parabola whose vertex = ( 2,-3) and foci = ( 0,5).
it is the form of x² = 4by.
⇒ vertex = ( 0,0)
⇒ foci = ( 0,b).
⇒ it s in the form of ( x - h)² = 4b( y - k).
⇒ ( x - 2 )² = 4(5)( y - (-3)).
⇒ ( x - 2 )² = 20 ( y + 3 ).
MORE INFORMATION.
EQUATION OF PARABOLA.
(1) y² = 4ax
conditions of tangent c = a/m.
equation of tangent y = mx + a/m.
point of contact = ( a/m² , 2a/m).
(2) = y² = -4ax.
conditions of tangent = c = -a/m.
equation of tangent = y = mx - a/m.
point of contact = ( -a/m² , 2a/m ).
(3) = x² = 4ay.
conditions of tangent = c = -am².
equation of tangent = y = mx - am².
point of contact = ( 2am,am²).
(4) = x² = 4ay.
conditions of tangent = c = am².
equation of tangent = y = mx + am².
point of contact = ( -2am,-am²).
Similar questions