The orthocentre of the triangle F1MN is [2016]

Question

Let $F_{1} (x_{1}, 0)$ and $F_{2} (x_{2}, 0)$ , for $x_{1} < 0$ and $x_{2} > 0$ , be the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1 .$ Suppose a parabola having vertex at the origin and focus at $F_{2}$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant. [2016]

Q. The orthocentre of the triangle $F_{1} M N$ is

Smart Achievers · Accepted Answer

(1)

$Given : Ellipse \frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$

$e = \sqrt{1 - \frac{8}{9}} = \frac{1}{3} \dots (i)$

$∴$ $F_{1} (- 1, 0) and F_{2} (1, 0)$

$Parabola with vertex at (0, 0) and focus at F_{2} (1, 0) is$

$y^{2} = 4 x \dots (i i)$

$∴$ $On solving (i) and (i i), we get the intersection points of ellipse and parabola as$

$M (\frac{3}{2}, \sqrt{6}) and N (\frac{3}{2}, - \sqrt{6})$

$One altitude of ∆ F_{1} M N is x-axis i.e. y = 0 and altitude from M to F_{1} N is$

$y - \sqrt{6} = \frac{5}{2 \sqrt{6}} (x - \frac{3}{2})$

$Putting y = 0 in above equation, we get$ $x = - \frac{9}{10}$

$∴$ $Orthocentre (- \frac{9}{10}, 0)$

Solution

1 $(- \frac{9}{10}, 0)$

2 $(\frac{2}{3}, 0)$

3 $(\frac{9}{10}, 0)$

4 $(\frac{2}{3}, \sqrt{6})$

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Solution

Q. Let F1(x1,0) and F2(x2,0), for x1<0 and x2>0, be the foci of the ellipse x29+y28=1. Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. [2016] Q. The orthocentre of the triangle F1MN is

1 (-910, 0)

2 (23, 0)

3 (910, 0)

4 (23, 6)

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1 $(- \frac{9}{10}, 0)$

2 $(\frac{2}{3}, 0)$

3 $(\frac{9}{10}, 0)$

4 $(\frac{2}{3}, \sqrt{6})$