JEE Challenger
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MathMedAdv 2016 • Paper 2

Orthocentre of Triangle Formed by Ellipse Parabola Intersection Points

Comprehension Passage

Let F1(x1,0)F_1(x_1, 0) and F2(x2,0)F_2(x_2, 0), where x1<0x_1 < 0 and x2>0x_2 > 0, denote the foci of an ellipse given by x29+y28=1\frac{x^2}{9} + \frac{y^2}{8} = 1. Consider a parabola with its vertex at the origin and focus at F2F_2 intersecting this ellipse at MM in the first quadrant and NN in the fourth quadrant.

Find the orthocentre of the triangle F1MNF_1 M N.

Options

A

(−910,0)\left(-\frac{9}{10}, 0\right)

Correct
B

(23,0)\left(\frac{2}{3}, 0\right)

C

(910,0)\left(\frac{9}{10}, 0\right)

D

(23,6)\left(\frac{2}{3}, \sqrt{6}\right)

Topics & Concepts

Conic SectionsEllipseParabola

Step-by-Step Solution

To find the orthocentre of the triangle F1MNF_1 M N, we first determine the coordinates of the foci of the given ellipse and the points of intersection MM and NN.

Step 1: Find the Foci of the Ellipse

The equation of the ellipse is given by: x29+y28=1\frac{x^2}{9} + \frac{y^2}{8} = 1

Here, a2=9a^2 = 9 and b2=8b^2 = 8. The eccentricity ee of the ellipse is: e=1−b2a2=1−89=13e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{8}{9}} = \frac{1}{3}

The foci are located at (±ae,0)=(±3⋅13,0)=(±1,0)(\pm ae, 0) = \left(\pm 3 \cdot \frac{1}{3}, 0\right) = (\pm 1, 0). Since x1<0x_1 < 0 and x2>0x_2 > 0, we have: F1=(−1,0)andF2=(1,0)F_1 = (-1, 0) \quad \text{and} \quad F_2 = (1, 0)

Step 2: Find the Points of Intersection MM and NN

The parabola has its vertex at the origin (0,0)(0, 0) and its focus at F2(1,0)F_2(1, 0). Therefore, the focal length is ap=1a_p = 1, and its equation is: y2=4xy^2 = 4x

Substitute y2=4xy^2 = 4x into the equation of the ellipse: x29+4x8=1  ⟹  x29+x2=1\frac{x^2}{9} + \frac{4x}{8} = 1 \implies \frac{x^2}{9} + \frac{x}{2} = 1

Multiplying the entire equation by 1818: 2x2+9x−18=02x^2 + 9x - 18 = 0

Factoring the quadratic equation: (2x−3)(x+6)=0(2x - 3)(x + 6) = 0

Since x>0x > 0 for real points on the parabola y2=4xy^2 = 4x, we take: x=32x = \frac{3}{2}

Corresponding yy-values: y2=4(32)=6  ⟹  y=±6y^2 = 4\left(\frac{3}{2}\right) = 6 \implies y = \pm \sqrt{6}

Thus, the points of intersection in the first and fourth quadrants are: M=(32,6)andN=(32,−6)M = \left(\frac{3}{2}, \sqrt{6}\right) \quad \text{and} \quad N = \left(\frac{3}{2}, -\sqrt{6}\right)

Step 3: Find the Orthocentre of △F1MN\triangle F_1 M N

The vertices of △F1MN\triangle F_1 M N are: F1(−1,0),M(32,6),N(32,−6)F_1(-1, 0), \quad M\left(\frac{3}{2}, \sqrt{6}\right), \quad N\left(\frac{3}{2}, -\sqrt{6}\right)

  1. First Altitude: Since MM and NN both have the same x-coordinate x=32x = \frac{3}{2}, the line segment MNMN is vertical. The altitude from F1(−1,0)F_1(-1, 0) to the side MNMN must be horizontal and passes through (−1,0)(-1, 0). Hence, the equation of this altitude is: y=0(the x-axis)y = 0 \quad \text{(the } x\text{-axis)}

    Therefore, the orthocentre H(xH,yH)H(x_H, y_H) lies on the xx-axis, meaning yH=0y_H = 0.

  2. Second Altitude: Now, consider the side F1NF_1 N. The slope of F1NF_1 N is: mF1N=−6−032−(−1)=−652=−265m_{F_1 N} = \frac{-\sqrt{6} - 0}{\frac{3}{2} - (-1)} = \frac{-\sqrt{6}}{\frac{5}{2}} = -\frac{2\sqrt{6}}{5}

    The altitude from vertex M(32,6)M\left(\frac{3}{2}, \sqrt{6}\right) is perpendicular to F1NF_1 N, so its slope m⊥m_\perp is: m⊥=−1mF1N=526m_\perp = -\frac{1}{m_{F_1 N}} = \frac{5}{2\sqrt{6}}

    The equation of the altitude from MM is: y−6=526(x−32)y - \sqrt{6} = \frac{5}{2\sqrt{6}} \left(x - \frac{3}{2}\right)

  3. Intersection of Altitudes: Since the orthocentre lies on y=0y = 0, substitute y=0y = 0 into the equation of the altitude from MM: 0−6=526(xH−32)0 - \sqrt{6} = \frac{5}{2\sqrt{6}} \left(x_H - \frac{3}{2}\right)

    −6⋅265=xH−32-\sqrt{6} \cdot \frac{2\sqrt{6}}{5} = x_H - \frac{3}{2}

    −125=xH−32  ⟹  xH=32−125=−910-\frac{12}{5} = x_H - \frac{3}{2} \implies x_H = \frac{3}{2} - \frac{12}{5} = -\frac{9}{10}

Thus, the orthocentre of triangle F1MNF_1 M N is: (−910,0)\left(-\frac{9}{10}, 0\right)

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