Normals are drawn at point P,Q,R lying on parabola y2=4x which intersect at (3,0) then area of △PQR is?

Solutions

Expert Solution

We know that standard equation of parabola is given as-

y² = 4ax

On differeniating we get

2y dy/dx=4

Slope of tangent =2/y

Slope of normal=−y/2

Equation of normal

y=(−y/2)(x−3)

y= (-xy -3y)/2

2y = -xy -3y

2y = -y(x+3)

2=3−x

2-3 = -x

-1=-x

x=1

y=0

Given that

Points P, Q, R are (0,0),(1,2) & (1,−2)

Base QR=4

Height =1

We know that

area of triangle =1/2 base x height

Thereforewe have,

Area =1/2 X 4 X1

=2.

Hence area of the traingle = 2.

Area of triangle = 2.

Nikhil Thakur answered 1 year ago

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