wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y22x4y=0 and the given parabola. The area of the triangle PQS is.

Open in App
Solution

Given: 'S' be gthe focus of the parabola y2=8x comparing given parabola with standard parabola y2=4ax.
4a=8 a=2
focus S(a,0)(2,0)
Now, PQ is the common chord of circle & parabola
point P must satisfy equation of parabola & circle.
as any point on parabola is P(at2,2at)
P(2t2,4t)......(a=2)
Equation of circle is x2+y22x4y=0
substituting, C-ordinates of P in rquation of circle.
i.e. x=2t2&y=4t
(2t2)2+(4t)2+2(2t2)4(4t)=0
4t4+16t24t216t=0
4t(t3+4tt4)=0
4t(t3+3t4)=0
t=0 and / or t3+3t4=0 t=1
Now, for t=0,P(0,0)
for t=1,Q(2,4)
[ Parabola & circle has common chord one point is P & another is Q]
To find Area of ΔPQS,
We have P(0,0),Q(2,4),S(2,0)

Method I :
Area =12×2×4
A(ΔPQS)=4units [Ref. image]

Method II :
Area =Px(QySy)+Qx(SyPy)+Sx(PyQy)2

=0(40)+2(00)+2(04)4
=0+082

Area=4 units

1170842_75_ans_582c093fccab451d86898e047a4903a4.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line and a Parabola
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon