JEE Challenger
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MathHardAdv 2016 • Paper 2

Ratio of Triangle to Quadrilateral Area in Ellipse Parabola Configuration

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.

Suppose tangents drawn to the ellipse at MM and NN intersect at RR, and the normal to the parabola at MM cuts the xx-axis at QQ. Determine the ratio of the area of triangle MQRMQR to the area of quadrilateral MF1NF2M F_1 N F_2.

Options

A

3:43:4

B

4:54:5

C

5:85:8

Correct
D

2:32:3

Topics & Concepts

Conic SectionsEllipseParabolaTangents and Normals

Step-by-Step Solution

To find the ratio of the area of triangle MQRMQR to the area of quadrilateral MF1NF2M F_1 N F_2, we determine the equations and coordinates of all relevant points step-by-step.

Step 1: Foci of the Ellipse

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

Here, a2=9  ⟹  a=3a^2 = 9 \implies a = 3 and b2=8  ⟹  b=22b^2 = 8 \implies b = 2\sqrt{2}.

The eccentricity ee is given by: 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)=(±1,0)(\pm ae, 0) = (\pm 1, 0). Since x1<0x_1 < 0 and x2>0x_2 > 0: F1=(−1,0)andF2=(1,0)F_1 = (-1, 0) \quad \text{and} \quad F_2 = (1, 0)

Step 2: Parabola and Intersection Points MM and NN

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

To find the intersection points MM and NN, we solve the system of equations for the ellipse and parabola: x29+4x8=1  ⟹  x29+x2=1\frac{x^2}{9} + \frac{4x}{8} = 1 \implies \frac{x^2}{9} + \frac{x}{2} = 1 2x2+9x−18=02x^2 + 9x - 18 = 0 (2x−3)(x+6)=0(2x - 3)(x + 6) = 0

Since x>0x > 0 for real intersection points: x=32x = \frac{3}{2}

Substituting x=32x = \frac{3}{2} into y2=4xy^2 = 4x: y2=4(32)=6  ⟹  y=±6y^2 = 4 \left(\frac{3}{2}\right) = 6 \implies y = \pm \sqrt{6}

Thus, the intersection points are: M=(32,6)(in the 1st quadrant)M = \left(\frac{3}{2}, \sqrt{6}\right) \quad (\text{in the 1st quadrant}) N=(32,−6)(in the 4th quadrant)N = \left(\frac{3}{2}, -\sqrt{6}\right) \quad (\text{in the 4th quadrant})

Step 3: Coordinates of Point RR

Point RR is the intersection of the tangents to the ellipse at MM and NN. By symmetry across the xx-axis, RR lies on the xx-axis (y=0y = 0).

The equation of the tangent to the ellipse at M(32,6)M\left(\frac{3}{2}, \sqrt{6}\right) is: x(32)9+y68=1  ⟹  x6+y68=1\frac{x \left(\frac{3}{2}\right)}{9} + \frac{y \sqrt{6}}{8} = 1 \implies \frac{x}{6} + \frac{y\sqrt{6}}{8} = 1

Setting y=0y = 0 gives: x6=1  ⟹  x=6\frac{x}{6} = 1 \implies x = 6

Thus, point RR is: R=(6,0)R = (6, 0)

Step 4: Coordinates of Point QQ

Point QQ is the point where the normal to the parabola at MM cuts the xx-axis.

Differentiating y2=4xy^2 = 4x with respect to xx: 2ydydx=4  ⟹  dydx=2y2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}

At M(32,6)M\left(\frac{3}{2}, \sqrt{6}\right), the slope of the tangent is: mt=26m_t = \frac{2}{\sqrt{6}}

The slope of the normal mnm_n is: mn=−62m_n = -\frac{\sqrt{6}}{2}

The equation of the normal line at MM is: y−6=−62(x−32)y - \sqrt{6} = -\frac{\sqrt{6}}{2}\left(x - \frac{3}{2}\right)

Setting y=0y = 0 to find the xx-intercept Q(xQ,0)Q(x_Q, 0): −6=−62(xQ−32)-\sqrt{6} = -\frac{\sqrt{6}}{2}\left(x_Q - \frac{3}{2}\right) 2=xQ−32  ⟹  xQ=722 = x_Q - \frac{3}{2} \implies x_Q = \frac{7}{2}

Thus, point QQ is: Q=(72,0)Q = \left(\frac{7}{2}, 0\right)

Step 5: Area of Triangle MQRMQR

The vertices of △MQR\triangle MQR are M(32,6)M\left(\frac{3}{2}, \sqrt{6}\right), Q(72,0)Q\left(\frac{7}{2}, 0\right), and R(6,0)R(6, 0).

Since QQ and RR lie on the xx-axis, the base QRQR is: Base QR=6−72=52\text{Base } QR = 6 - \frac{7}{2} = \frac{5}{2}

The height is the yy-coordinate of point MM: Height=6\text{Height} = \sqrt{6}

Area(△MQR)=12×Base×Height=12×52×6=564\text{Area}(\triangle MQR) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times \frac{5}{2} \times \sqrt{6} = \frac{5\sqrt{6}}{4}

Step 6: Area of Quadrilateral MF1NF2M F_1 N F_2

The vertices of quadrilateral MF1NF2M F_1 N F_2 are: M(32,6),F1(−1,0),N(32,−6),F2(1,0)M\left(\frac{3}{2}, \sqrt{6}\right), \quad F_1(-1, 0), \quad N\left(\frac{3}{2}, -\sqrt{6}\right), \quad F_2(1, 0)

Since focus F2(1,0)F_2(1,0) lies inside △MF1N\triangle MF_1N, quadrilateral MF1NF2M F_1 N F_2 is non-convex (dart-shaped). Its area is the difference between the areas of △MF1N\triangle MF_1N and △MF2N\triangle MF_2N:

Area(MF1NF2)=Area(△MF1N)−Area(△MF2N)\text{Area}(M F_1 N F_2) = \text{Area}(\triangle MF_1N) - \text{Area}(\triangle MF_2N)

Both triangles share the vertical base MN=26MN = 2\sqrt{6}:

  • Height of △MF1N=32−(−1)=52\triangle MF_1N = \frac{3}{2} - (-1) = \frac{5}{2}
  • Height of △MF2N=32−1=12\triangle MF_2N = \frac{3}{2} - 1 = \frac{1}{2}

Area(△MF1N)=12×26×52=562\text{Area}(\triangle MF_1N) = \frac{1}{2} \times 2\sqrt{6} \times \frac{5}{2} = \frac{5\sqrt{6}}{2} Area(△MF2N)=12×26×12=62\text{Area}(\triangle MF_2N) = \frac{1}{2} \times 2\sqrt{6} \times \frac{1}{2} = \frac{\sqrt{6}}{2}

Area(MF1NF2)=562−62=26\text{Area}(M F_1 N F_2) = \frac{5\sqrt{6}}{2} - \frac{\sqrt{6}}{2} = 2\sqrt{6}

Step 7: Ratio of Areas

Ratio=Area(△MQR)Area(MF1NF2)=56426=58\text{Ratio} = \frac{\text{Area}(\triangle MQR)}{\text{Area}(M F_1 N F_2)} = \frac{\frac{5\sqrt{6}}{4}}{2\sqrt{6}} = \frac{5}{8}

Hence, the required ratio is 5:85:8, which corresponds to Option C.

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