If vertex is at (1, 2) and focus (2, 0) then find the equation of the parabola.(a) y^2-8x+4y+12=0(b) y^2-8x-4y-12=0(c) y^2-8x-4y+12=0(d) y^2+8x+4y+12=0The question was asked in final exam.My enquiry is from Conic Sections in chapter Conic Sections of Mathematics – Class 11

If vertex is at (1, 2) and focus (2, 0) then find the equation of the

1.	If vertex is at (1, 2) and focus (2, 0) then find the equation of the parabola.(a) y^2-8x+4y+12=0(b) y^2-8x-4y-12=0(c) y^2-8x-4y+12=0(d) y^2+8x+4y+12=0The question was asked in final exam.My enquiry is from Conic Sections in chapter Conic Sections of Mathematics – Class 11
Answer» The CORRECT option is (C) y^2-8x-4y+12=0 Easiest EXPLANATION: Since vertex is at (1, 2) and FOCUS (2, 0) so parabola equation will be (y-2)^2=4*2(x-1) y^2-8x-4y+12=0.

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If vertex is at (1, 2) and focus (2, 0) then find the equation of the parabola.(a) y^2-8x+4y+12=0(b) y^2-8x-4y-12=0(c) y^2-8x-4y+12=0(d) y^2+8x+4y+12=0The question was asked in final exam.My enquiry is from Conic Sections in chapter Conic Sections of Mathematics – Class 11

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