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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). |
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Answer» Correct Answer - `y=tanalpha(x-a)` 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. |
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