Let `L_(1),L_(2)andL_(3)` be the three normals to the parabola `y^(2)=4ax` from point P inclined at the angle `theta_(1),theta_(2)andtheta_(3)` with x-axis, respectively. Then find the locus of point P given that `theta_(1)+theta_(2)+theta_(3)=alpha` (constant).

Answer

`y=tanalpha(x-a)`

Explanation

Equation of normal to parabola `y^(2)=4ax` having slope m is
`y=mx-2am-am^(3)`
This normal passes through point P(h,k)
`:." "k=mh-2am-am^(3)`
`or" "am^(3)+(2a-h)m+k=0` (1)
This equation has three real roots `m_(1),m_(2)andm_(3)`, which are slopes of three normals.
Given that `m_(1)=tantheta_(1),m_(2)=tantheta_(2)andm_(3)=tantheta_(3)`
From equation (1), we have
`m_(1)+m_(2)+m_(3)=0,`
`m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1)=(2a-h)/(a),`
`and" "m_(1)m_(2)m_(3)=(-k)/(a)`
Now, `theta_(1)+theta_(2)+theta_(3)=alpha`
`:." "tan(theta_(1)+theta_(2)+theta_(3))=tanalpha`
`rArr" "(m_(1)+m_(2)+m_(3)-m_(1)m_(2)m_(3))/(1-m_(1)m_(2)-m_(2)m_(3)-m_(3)m_(1))=tanalpha`
`rArr" "(0+((k)/(a)))/(1-((2a-h)/(a)))=tanalpha`
`rArr" "y=tanalpha(x-a)`,
which is the required locus.