Logo
Jeeify
MCQMediumJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2025 Question with Solution

Let the parabola y=x2+px−3y = x^2 + px - 3 meet the coordinate axes at the points P, Q and R. If the circle C with centre at (−1,−1)(-1, -1) passes through the points P, Q and R, then the area of △PQR\triangle PQR is:

  • A

    44

  • B

    66

  • C

    77

  • D

    55

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The parabola is y=x2+px−3y = x^2 + px - 3 and the circle has centre (−1,−1)(-1,-1). The parabola meets the axes at points P, Q and R.

Find: The area of △PQR\triangle PQR.

The intersections with the xx-axis are the roots of

x2+px−3=0x^2 + px - 3 = 0

Let these roots be r1r_1 and r2r_2. Then

P(r1,0),Q(r2,0)P(r_1,0), \qquad Q(r_2,0)

The yy-intercept is obtained by putting x=0x=0, so

y=−3y = -3

Hence

R(0,−3)R(0,-3)

Since the circle with centre (−1,−1)(-1,-1) passes through R(0,−3)R(0,-3), its radius satisfies

r2=(0+1)2+(−3+1)2=1+4=5r^2 = (0+1)^2 + (-3+1)^2 = 1 + 4 = 5

So the equation of the circle is

(x+1)2+(y+1)2=5(x+1)^2 + (y+1)^2 = 5

Now for any xx-intercept of the parabola, the point is of the form (r,0)(r,0). Substituting in the circle,

(r+1)2+(0+1)2=5(r+1)^2 + (0+1)^2 = 5 (r+1)2+1=5(r+1)^2 + 1 = 5 (r+1)2=4(r+1)^2 = 4

Thus

r+1=±2r+1 = \pm 2

so

r=1orr=−3r = 1 \quad \text{or} \quad r = -3

Therefore,

P(1,0),Q(−3,0),R(0,−3)P(1,0), \qquad Q(-3,0), \qquad R(0,-3)

Take PQPQ as the base. Its length is

∣1−(−3)∣=4|1-(-3)| = 4

The perpendicular distance of R(0,−3)R(0,-3) from the xx-axis is

33

Hence the area is

Area=12×4×3=6\text{Area} = \frac{1}{2} \times 4 \times 3 = 6

Therefore, the area of △PQR\triangle PQR is 66. The correct option is B.

Using base and height directly

Given: RR is the yy-intercept of the parabola, so R(0,−3)R(0,-3). The circle has centre (−1,−1)(-1,-1) and passes through all three intercept points.

Find: The area of △PQR\triangle PQR.

From the circle, any point on the xx-axis lying on it has the form (r,0)(r,0). So

(r+1)2+1=5(r+1)^2 + 1 = 5 (r+1)2=4(r+1)^2 = 4

which gives the two xx-intercepts directly as

r=1,−3r = 1, -3

Thus the base on the xx-axis is

PQ=4PQ = 4

and the height from R(0,−3)R(0,-3) to the xx-axis is

33

Therefore,

Area=12×4×3=6\text{Area} = \frac{1}{2} \times 4 \times 3 = 6

So the correct option is B.

Common mistakes

  • A common mistake is to use only the parabola and try to determine pp first. That is unnecessary here because the circle condition directly gives the xx-intercepts. Use the circle equation at points of the form (r,0)(r,0) instead.

  • Students may take the height as −3-3 because point RR has coordinate y=−3y=-3. Height is a distance, so it must be positive. Use height =3=3.

  • Another mistake is to forget that the base length is the distance between the two xx-intercepts, namely ∣1−(−3)∣|1-(-3)|, not 1+(−3)1+(-3). Always take absolute difference for length.

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Conic Sections (Parabola, Ellipse, Hyperbola)

Related questions