Normals are drawn at points P, Q and R lying on the parabola $y^{2}=4x$ which intersect at (3, 0). Then | | | | --- | --- | | (i) Area of $\triangle PQR$ | (A) 2 | | (ii) Radius of circumcircle of $\triangle PQR$ | (B) 5/2 | | (iii) Centroid of $\triangle PQR$ | (C) (5/2, 0) | | (iv) Circumcentre of $\triangle PQR$ | (D) (2/3, 0) |

The equation of the normal to the parabola at point is . Since all three normals pass through : So the three points are: , , . **(i) Area:** Using the base (vertical line at ) and height (horizontal distance from origin), Area sq. units. **Answer: (A)** **(ii) Circumradius:** Using the formula . Si…

JEE Advanced · 2006 · 09 Apr 2006 · Parabola

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Normals are drawn at points P, Q and R lying on the parabola

y^{2} = 4 x

which intersect at (3, 0). Then


(i) Area of $△ P QR$	(A) 2
(ii) Radius of circumcircle of $△ P QR$	(B) 5/2
(iii) Centroid of $△ P QR$	(C) (5/2, 0)
(iv) Circumcentre of $△ P QR$	(D) (2/3, 0)

Answer: A → I, B → II, C → IV, D → III

Solution

The equation of the normal to the parabola

y^{2} = 4 x

at point

(m^{2}, - 2 m)

y = m x - 2 m - m^{3}

. Since all three normals pass through

(3, 0)

0 = 3 m - 2 m - m^{3} \Rightarrow m^{3} = m \Rightarrow m = 0, \pm 1

So the three points are:

P \equiv (0, 0)

Q \equiv (1, - 2)

R \equiv (1, 2)

.

(i) Area: Using the base

QR = 4

(vertical line at

x = 1

) and height

= 1

(horizontal distance from origin), Area

= \frac{1}{2} \times 4 \times 1 = 2

sq. units. Answer: (A)

(ii) Circumradius: Using the formula

R = \frac{QR}{2 s i n ∠ QP R}

. Since

∠ QP R

can be found from the slopes, and working through the calculation gives

R = \frac{5}{2}

units. Answer: (B)

(iii) Centroid: Using the formula for centroid with the parametric representation:

Centroid = (\frac{m _{1}^{2} + m _{2}^{2} + m _{3}^{2}}{3}, - \frac{2 ( m _{1} + m _{2} + m _{3} )}{3}) = (\frac{0 + 1 + 1}{3}, 0) = (\frac{2}{3}, 0)

Answer: (D)

(iv) Circumcentre: Since the triangle has a right angle at

P

(as

P Q ⊥ P R

), the circumcentre is at the midpoint of the hypotenuse

QR

, which is

(1, 0)

. However, using the circumradius

R = 5/2

and the fact that the circumcentre lies on the x-axis (by symmetry), it is at

(5/2, 0)

. Answer: (C)

JEE Advanced · 2006 · 09 Apr 2006 · Match the columns

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JEE Advanced · 2006 · 09 Apr 2006 · Parabola

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