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Question

Find the equation of the ellipse in the following cases:
(i) focus is (0, 1), directrix is x + y = 0 and e = 12
(ii) focus is (−1, 1), directrix is x − y + 3 = 0 and e = 12
(iii) focus is (−2, 3), directrix is 2x + 3y + 4 = 0 and e = 45
(iv) focus is (1, 2), directrix is 3x + 4y − 5 = 0 and e = 12.

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Solution

(i)



Let S(0,1) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have:SP=e×PM⇒SP=12×PM⇒2SP=PM⇒4SP2=PM2⇒4x2+y-12=x+y12+122⇒4x2+y2+1-2y=x2+y2+2xy2⇒8x2+8y2+8-16y=x2+y2+2xy⇒7x2+7y2-2xy-16y+8=0This is the required equation of the ellipse.

(ii)

Let S(-1,1) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have:SP=e×PM⇒SP=12×PM⇒2SP=PM⇒4SP2=PM2⇒4x+12+y-12=x-y+312+-122⇒4x2+1+2x+y2+1-2y=x2+y2+9-2xy-6y+6x2⇒8x2+8+16x+8y2+8-16y=x2+y2+9-2xy-6y+6x⇒7x2+7y2+2xy-10y+10x+7=0This is the required equation of the ellipse.

(iii)


Let S(-2,3) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have: SP=e×PM⇒SP=45×PM⇒54SP=PM⇒2516SP2=PM2⇒2516x+22+y-32=2x+3y+422+322⇒2516x2+4+4x+y2+9-6y=4x2+9y2+16+12xy+24y+16x13⇒325x2+4+4x+y2+9-6y=164x2+9y2+16+12xy+24y+16x⇒325x2+1300+1300x+325y2+2925-1950y=64x2+144y2+256+192xy+384y+256x⇒261x2+181y2+1044x-2309y-192xy+3969=0This is the required equation of the ellipse.

(iv)

Let S(1,2) be the focus and ZZ' be the directrix.Let P(x,y) be any point on the ellipse and let PM be the perpendicular from P on the directrix.Then by the definition, we have:SP=e×PM⇒SP=12×PM⇒2SP=PM⇒4SP2=PM2⇒4x-12+y-22=3x+4y-532+422⇒4x2+1-2x+y2+4-4y=9x2+16y2+25+24xy-40y+30x25⇒100x2+100-200x+100y2+400-400y=9x2+16y2+25+24xy-40y-30x⇒91x2+84y2-24xy-360y-170x+475=0This is the required equation of the ellipse.

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