Let the parabola y=x^2+px-3 meet the coordina | vedclass

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(B) The parabola is given by $y=x^2+px-3$.The points of intersection with the coordinate axes are $P, Q$ and $R$.The $y$-intercept is found by setting

Vedclass · Accepted Answer

$6$

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(B) The parabola is given by $y=x^2+px-3$.The points of intersection with the coordinate axes are $P, Q$ and $R$.The $y$-intercept is found by setting $x=0$,which gives $y=-3$. Thus,$R=(0,-3)$.The $x$-intercepts are found by setting $y=0$,giving $x^2+px-3=0$. Let the roots be $\alpha$ and $\beta$,so $P=(\alpha, 0)$ and $Q=(\beta, 0)$.The circle with center $(-1,-1)$ has the equation $(x+1)^2+(y+1)^2=r^2$.Since the circle passes through $R(0,-3)$,we have $(0+1)^2+(-3+1)^2=r^2$,which gives $1^2+(-2)^2=r^2$,so $r^2=5$.The equation of the circle is $(x+1)^2+(y+1)^2=5$.To find the $x$-intercepts of the circle,set $y=0$: $(x+1)^2+(0+1)^2=5 \implies (x+1)^2=4 \implies x+1=\pm 2$.Thus,$x=1$ or $x=-3$. So $P=(1,0)$ and $Q=(-3,0)$.The area of $	riangle PQR$ with vertices $(1,0), (-3,0)$ and $(0,-3)$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Base $PQ = |1 - (-3)| = 4$.Height (distance of $R$ from $x$-axis) $= |-3| = 3$.Area $= \frac{1}{2} 	imes 4 	imes 3 = 6$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

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

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