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Let F_(1)(x_(1),0)and F_(2)(x_(2),0)" for "x_(1)lt0 and x_(2)gt0 the foci of the ellipse (x^(2))/(9)+(y^(2))/(8)=1. Suppose a parabola having vertex at the origin and focus at F_(2) intersects the ellipse at point M in the first quadrant and ata point N in the fourth quardant. The orthocentre of the triangle F_(1)MN, is |
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Answer» `(-(9)/(10),0)` `therefore e-sqrt(1-(b^(2))/(a^(2)))-sqrt(1-(8)/(9))=(1)/(3)`. So, the coordinates of foci are `F_(1)(-1,0)` and `F_(2)(1, 0)`. Theequation of the PARABOLA having vertex at theorigin and focusat `F_(2)(1, 0) is y^(2)=4x`. Solving `(x^(2))/(9)+(y^(2))/(8)=1 and y^(2)=4x`, we find that the coordinates of M and N are `((3)/(2),sqrt6) and ((3)/(2),-sqrt6)` respectively. Clearly, ALTITUDE through `F_(1) " of " F_(1)MN` is a x-axis i.e. y = 0. the equation of the altitude through `M((3)/(2),sqrt6)`is `y-sqrt6=(5)/(2sqrt6)(x-(3)/(2))` Solving the equation with y = 0, we obtain `(-9//10, 0)` the orthocentre of `DeltaF_(1)MN`. |
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