Please solve Q3 and please solve it full Q3 Find the equation of the parabola that satisfies the - Maths - Conic Sections

Question

Please solve Q3 and please solve it full

Q3 Find the equation of the parabola that satisfies the (i) Focus
(6, 0) ; directrix x = – 6 (ii) Focus (0,–3); directrix
y = 3 (iii) vertex (0,0) ; focus (3, 0) (iv) vertex (0, 0) ; focus
(–2, 0) (v) vertex (0,0), Passing through (2, 3) and axis is
along x-axis (vi) vertex (0, 0), passing through (5, 2) and
symmetric with respect to y-axis

Muskaan Talwar · Accepted Answer

Dear Student,

i Since focus 6,0 lies on +ve x-axis and directrix is x=-6thus, parabola is of form y2=4ax with a=6So required equation is y2=24xii Since focus 0,-3 lies on -ve y-axis and directrix is y=3thus, parabola is of form x2=-4ay with a=3So required equation is x2=-12yiii Since focus 3,0 lies on +ve x-axis and vertex is 0,0thus, parabola is of form y2=4ax with a=3So required equation is y2=12xiv Since focus -2,0 lies on -ve x-axis and vertex is 0,0thus, parabola is of form y2=-4ax with a=2So required equation is y2=-8xv Since vertex is 0,0 and axis is along x-axis and passes through 2,3so it passes through 1st quadrant,thus, parabola is of form y2=4ax Now, 32=4a2⇒9=8a⇒a=98Thus required equation is y2=498x=92xvi Since vertex is 0,0 and axis is along y-axis and passes through 5,2so it passes through 1st quadrant,thus, parabola is of form x2=4ay Now, 52=4a2⇒25=8a⇒a=258Thus required equation is x2=4258y=252y

Regards