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Consider the hyperbola (X^(2))/(9)-(y^(2))/(a^(2))=1 and the circle x^(2)+(y-3)=9. Also, the given hyperbola and the ellipse (x^(2))/(41)+(y^(2))/(16)=1 are orthogonal to each other. A variable line cuts the circle at point A and B and it cuts the hyperbola atpoints C and D. The locus of midpoint of AB such that tangents at points C and D always intersect each other at the directrix of the hyperbola, is |
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Answer» `X^(2)+y^(2)pm5x-3y=0` `"So,"a_(h)e_(h)=a_(e)e_(e)` `rArr""a^(2)+9=41-16` `rArr""9+a^(2)=25` `rArr""a^(2)=16` Thus, hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1.` So, common tangents to the circle and hyperbola are `x = pm 3`. Director circle of hyperbola does not EXIST as `a LT b.` Director circle of circle is `x^(2)+(y-3)^(2)=18` `rArr""x^(2)+y^(2)-6y-9=0` This meets the hyperbola `16x^(2)-9y^(2)=144` at four points from where tangents drawn to the circle `x^(2)+(y-3)^(2)=9` are perpendicular to each other. Let midpoint of AB be (h,k). So, equation of LINE AB is `hx+ky-3(y+k)=h^(2)+k^(2)-6k`. Since tangents at C and D intersect at the directrix, CD is the focal chord of hyperbola. So, AB passes through focus of the hyperbola and that is `(pm5,0)`. Therefore, required lacus is `x^(2)+y^(2)pm5x-3y=0`. |
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