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Question

Find the equation of the parabola whose focus is the point (2,3) and directrix is the line x-4y+3=0.Also,find the length of its latus-rectum.

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Solution

Let P(x,y) be any point on the parabola whose focus is S(2,3) and the directirx x-4y+3=0.

Draw PM perpendicular from P(x,y) on the directrix x-4y+3=0.

Then by definition SP=PM

SP2=PM2(x2)2+(y3)2=(x4y+3(1)2+(4)2)2x2+44x+y2+96y=((x4y+3)2(17)2))x2+y24x6y+4+9=(x4y+3)21717(x2+y24x6y+13)=(x4y+3)217x2+17y268x102y+221=x2+(4y)2+32+2×3×x17x2+17y268x102y+221=x2+16y2+98xy24y+6x17x2x2+17y216y2+8xy68x6x102y+24y+2219=016x2+y2+8xy74x78y+212=0

This is the equation of the required parabola.

Latus Rectum=Length of perpendicular from focus (2,3) on directrix x-4y+3=0

=2212+31+16=2717=1417


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