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.
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⇒(x−2)2+(y−3)2=(x−4y+3√(1)2+(−4)2)2⇒x2+4−4x+y2+9−6y=((x−4y+3)2√(17)2))⇒x2+y2−4x−6y+4+9=(x−4y+3)217⇒17(x2+y2−4x−6y+13)=(x−4y+3)2⇒17x2+17y2−68x−102y+221=x2+(−4y)2+32+2×3×x⇒17x2+17y2−68x−102y+221=x2+16y2+9−8xy−24y+6x⇒17x2−x2+17y2−16y2+8xy−68x−6x−102y+24y+221−9=0⇒16x2+y2+8xy−74x−78y+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
=2∣∣∣2−12+3√1+16∣∣∣=2∣∣∣−7√17∣∣∣=14√17