JEE Challenger
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MathMedAdv 2011 • Paper 1

Ratio of Triangle Areas Formed by Parabola Latus Rectum and Tangents

Consider the parabola y2=8xy^2 = 8x. Let Δ1\Delta_1 be the area of the triangle formed by the end points of its latus rectum and the point P(12,2)P\left(\frac{1}{2}, 2\right) on the parabola, and Δ2\Delta_2 be the area of the triangle formed by drawing tangents at PP and at the end points of the latus rectum. Then Δ1Δ2\frac{\Delta_1}{\Delta_2} is

Official Numerical Answer2

Topics & Concepts

Conic SectionsParabolaTangents and Normals

Step-by-Step Solution

To find the ratio Δ1Δ2\frac{\Delta_1}{\Delta_2}, we first determine the coordinates of the vertices for both triangles.

The equation of the parabola is:
y2=8x=4ax  ⟹  a=2y^2 = 8x = 4ax \implies a = 2

1. Calculation of Δ1\Delta_1

The endpoints of the latus rectum for the parabola y2=4axy^2 = 4ax are L(a,2a)L(a, 2a) and L′(a,−2a)L'(a, -2a).
Substituting a=2a = 2, we get:
L=(2,4),L′=(2,−4)L = (2, 4), \quad L' = (2, -4)
The third point is given as:
P=(12,2)P = \left(\frac{1}{2}, 2\right)

The area Δ1\Delta_1 of △PLL′\triangle P L L' is given by:
Δ1=12∣xP(yL−yL′)+xL(yL′−yP)+xL′(yP−yL)∣\Delta_1 = \frac{1}{2} \left| x_P(y_L - y_{L'}) + x_L(y_{L'} - y_P) + x_{L'}(y_P - y_L) \right|
Δ1=12∣12(4−(−4))+2(−4−2)+2(2−4)∣\Delta_1 = \frac{1}{2} \left| \frac{1}{2}(4 - (-4)) + 2(-4 - 2) + 2(2 - 4) \right|
Δ1=12∣12(8)+2(−6)+2(−2)∣\Delta_1 = \frac{1}{2} \left| \frac{1}{2}(8) + 2(-6) + 2(-2) \right|
Δ1=12∣4−12−4∣=12∣−12∣=6\Delta_1 = \frac{1}{2} |4 - 12 - 4| = \frac{1}{2} |-12| = 6

2. Calculation of Δ2\Delta_2

Let the three points on the parabola y2=4axy^2 = 4ax in parametric form (at2,2at)(at^2, 2at) be:

  • For L(2,4)L(2, 4): 2at1=4  ⟹  2(2)t1=4  ⟹  t1=12at_1 = 4 \implies 2(2)t_1 = 4 \implies t_1 = 1
  • For L′(2,−4)L'(2, -4): 2at2=−4  ⟹  2(2)t2=−4  ⟹  t2=−12at_2 = -4 \implies 2(2)t_2 = -4 \implies t_2 = -1
  • For P(12,2)P\left(\frac{1}{2}, 2\right): 2at3=2  ⟹  2(2)t3=2  ⟹  t3=122at_3 = 2 \implies 2(2)t_3 = 2 \implies t_3 = \frac{1}{2}

The intersection of the tangents at points with parameters tit_i and tjt_j on the parabola y2=4axy^2 = 4ax is given by:
(x,y)=(atitj,a(ti+tj))(x, y) = (a t_i t_j, a(t_i + t_j))

The vertices of the triangle formed by the three tangents are:

  1. Intersection of tangents at L(t1=1)L(t_1 = 1) and L′(t2=−1)L'(t_2 = -1):
    A=(2(1)(−1),2(1−1))=(−2,0)A = (2(1)(-1), 2(1 - 1)) = (-2, 0)

  2. Intersection of tangents at L′(t2=−1)L'(t_2 = -1) and P(t3=12)P\left(t_3 = \frac{1}{2}\right):
    B=(2(−1)(12),2(−1+12))=(−1,−1)B = \left(2(-1)\left(\frac{1}{2}\right), 2\left(-1 + \frac{1}{2}\right)\right) = (-1, -1)

  3. Intersection of tangents at P(t3=12)P\left(t_3 = \frac{1}{2}\right) and L(t1=1)L(t_1 = 1):
    C=(2(12)(1),2(12+1))=(1,3)C = \left(2\left(\frac{1}{2}\right)(1), 2\left(\frac{1}{2} + 1\right)\right) = (1, 3)

Now, the area Δ2\Delta_2 of △ABC\triangle ABC is:
Δ2=12∣xA(yB−yC)+xB(yC−yA)+xC(yA−yB)∣\Delta_2 = \frac{1}{2} \left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|
Δ2=12∣−2(−1−3)+(−1)(3−0)+1(0−(−1))∣\Delta_2 = \frac{1}{2} \left| -2(-1 - 3) + (-1)(3 - 0) + 1(0 - (-1)) \right|
Δ2=12∣−2(−4)−3+1∣\Delta_2 = \frac{1}{2} \left| -2(-4) - 3 + 1 \right|
Δ2=12∣8−3+1∣=12∣6∣=3\Delta_2 = \frac{1}{2} |8 - 3 + 1| = \frac{1}{2} |6| = 3

(Alternatively, by a standard property of parabolas, the area of a triangle formed by three points on a parabola is always twice the area of the triangle formed by the tangents drawn at those three points, i.e., Δ1=2Δ2\Delta_1 = 2\Delta_2.)

3. Conclusion

Δ1Δ2=63=2\frac{\Delta_1}{\Delta_2} = \frac{6}{3} = 2

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