19first curve is a parabola
eccentricity of S is 1/√2
.....now its manipulation and some calculations !
Multi answer
Q. Consider a curve 13[(x-2)2+(y-3)2] = (2x+3y-5)2. Tangents are drawn from (1,1) to this curve. If a conic S=0 having eccentricity half the eccentricity of the conic xy+x+y=0 always touches above tangents then the eccentricity of the locus of the centre of the conic S=0 is always less than
(a) 0.5 (b) 0.33 (c) 1 (d) 2
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5 Answers
19
1i think it's not that easy
check this
as organic pointed this is parabola with vertex (6/7 , 12/7)
eccentricity of S=1/√2
now (1,1) is a point on the directrix of the parabola so the tangents drawn to parabola are perpendicular and the chord of contact is it's focal chord!!!!
now since the eccentricity is less than 1 the curve should be ellipse
acc. to question the tangents from (1,1) are also the tangents to this ellipse and since the tangents are perpendicular the locus is the director circle!!!!!
now use these and try to get the answer..........
1and also i have got some part of eq. of ellipse
\frac{(x-h)^{2}}{2/3} + \frac{(y-k)^{2}}{1/3} =1
(h,k) centre