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

Correct answer: C. 6 y=x^2+px-3 Let P(α, 0), Q(β, 0), R(0,-3) Circle with centre (-1,-1) is (x+1)^2+(y+1)^2=r^2 Passes through (0,-3) aligned & .1^2+(-2)^2=r^2] & r^2=5 aligned (x+1)^2+(y+1)^2=5 Put y=0 aligned & (x+1)^2=5-1 & (x+1)^2=4 & x+1= ± 2 & x=1 or x=-3 & P(1,0) and Q(-3,0) aligned Area of PQR=(1)/(2)| array ccc 1 & 0 & 1 -3 & 0 & 1 0 & -3 & 1 array |=6

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Parabola

Let the parabola y=x^2+p x-3 , meet the coordinate axes at the...

Let the parabola

y=x^2+\mathrm{p} x-3

, meet the coordinate axes at the points

\mathrm{P}, \mathrm{Q}

and R . If the circle C with centre at

(-1,-1)

passes through the points

P, Q

and

R

, then the area of

\triangle P Q R

is :

Options

A $7$
B $4$
C $6$
D $5$

Correct answer

C. $6$

Step-by-step solution

\mathrm{y}=\mathrm{x}^2+\mathrm{px}-3

Let

\mathrm{P}(\alpha, 0), \mathrm{Q}(\beta, 0), \mathrm{R}(0,-3)

Circle with centre

(-1,-1)

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

Passes through

(0,-3)

\begin{aligned} & \left.1^2+(-2)^2=r^2\right] \\ & \mathrm{r}^2=5 \end{aligned}

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

Put

\mathrm{y}=0

\begin{aligned} & (x+1)^2=5-1 \\ & (x+1)^2=4 \\ & x+1= \pm 2 \\ & x=1 \text{\text{ or } } x=-3 \\ & \therefore P(1,0) \text{ and } Q(-3,0) \end{aligned}

Area of $\triangle \mathrm{PQR}=\frac{1}{2}\left|

\begin{array}{ccc}1 & 0 & 1 \\ -3 & 0 & 1 \\ 0 & -3 & 1\end{array}$$$\right|=6

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