Question

Find the locus of a point \[O\] when the three normals drawn from it are such that the sum of the three angles made by them with the axis is constant.

Answer 1

Hint: Take the directrix as \[x=-a\]and find the normal of the tangent by taking the slope as \['m'\]and take the angles as \[{{\theta }_{1}},{{\theta }_{2}}\]and \[{{\theta }_{3}}\]respectively.

Complete step-by-step answer:

Consider the figure that is drawn below

From the figure, it shows that the tangents at the extremities of any focal chord intersect at right angles in the directrix.

Now, the equation of the tangent at \[\left( {{x}_{1}},{{y}_{1}} \right)\Rightarrow y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)\]

The tangent’s in term of its slope \['m'\Rightarrow y=mx+\dfrac{a}{m}\]at \[\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)\]

The normal will be \[y-{{y}_{1}}=\dfrac{-{{y}_{1}}}{2a}\left( x-{{x}_{1}} \right)\]at \[\left( {{x}_{1}},{{y}_{1}} \right)\].

\[\therefore \]Normal \[\Rightarrow y=mx-2am-a{{m}^{3}}\]at \[\left( a{{m}^{2}},-2am \right)\].

Let the point \[O\]be taken as \[\left( h,k \right)\]

We know the equation of normal in slope form is

\[y=mx-2am-a{{m}^{3}}......\left( i \right)\]

It passes through the points \[O\left( h,k \right)\]

Put \[x=h\]and \[y=k\]

\[k=mh-2am-a{{m}^{3}}\]

Rearranging the equation, we get,

\[a{{m}^{3}}+2am-mh+k=0\]

\[\Rightarrow a{{m}^{3}}+m\left( 2a-h \right)+k=0.....\left( ii \right)\]

Let the three normals be drawn and their slopes are \[{{m}_{1}},{{m}_{2}}\]and \[{{m}_{3}}\].

In general, three normals can be drawn from any point to a parabola and the algebraic sum of the ordinates of these normals is zero.

For cubic, the sum of slopes of the normals is zero.

i.e. \[{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0.....\left( iii \right)\]

The perpendicular to the normal is the product of the root of slopes which will be \[\left( \dfrac{-k}{a} \right)\].

i.e. \[{{m}_{1}}.{{m}_{2}}.{{m}_{3}}=\dfrac{-k}{a}.....\left( iv \right)\]

Let \[{{\theta }_{1}},{{\theta }_{2}}\]and \[{{\theta }_{3}}\]be the angles made by them with the axis of the parabola.

Given that \[{{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}}=\text{constant}\].

Let us consider that

\[\tan \left( {{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}} \right)=\tan \left( \text{constant} \right)\]

By expanding \[\tan \left( {{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}} \right)\], we get

\[\dfrac{\tan {{\theta }_{1}}+\tan {{\theta }_{2}}+\tan {{\theta }_{3}}-\tan {{\theta }_{1}}.\tan {{\theta }_{2}}.\tan {{\theta }_{3}}}{1-\tan {{\theta }_{1}}\tan {{\theta }_{2}}-\tan {{\theta }_{2}}\tan {{\theta }_{3}}-\tan {{\theta }_{3}}\tan {{\theta }_{1}}}=\tan c.....\left( v \right)\]

where \[c\]is the constant.

\[\therefore \tan c\approx c\text{ }[\tan c\text{ can be approximated to }c]\]

\[{{m}_{1}}{{m}_{2}}+{{m}_{2}}{{m}_{3}}+{{m}_{3}}{{m}_{1}}=\dfrac{2a-h}{a}.....\left( vi \right)\]

Equation \[\left( vi \right)\]can be considered from the equation of tangents corresponding to three normals.

Equation \[\left( v \right)\]can be written as

\[\dfrac{\tan {{\theta }_{1}}+\tan {{\theta }_{2}}+\tan {{\theta }_{3}}-\tan {{\theta }_{1}}.\tan {{\theta }_{2}}.\tan {{\theta }_{3}}}{1-\left( \tan {{\theta }_{1}}\tan {{\theta }_{2}}+\tan {{\theta }_{2}}\tan {{\theta }_{3}}+\tan {{\theta }_{3}}\tan {{\theta }_{1}} \right)}=c.....\left( vii \right)\]

\[\left( \tan {{\theta }_{1}}+\tan {{\theta }_{2}}+\tan {{\theta }_{3}} \right)\]is equal to \[\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)=0\]

\[\tan {{\theta }_{1}}+\tan {{\theta }_{2}}+\tan {{\theta }_{3}}=0....\left( viii \right)\]

\[\tan {{\theta }_{1}}.\tan {{\theta }_{2}}.\tan {{\theta }_{3}}=\dfrac{-k}{a}\]which is similar to

\[{{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}....\left( ix \right)\]

\[\tan {{\theta }_{1}}\tan {{\theta }_{2}}+\tan {{\theta }_{2}}\tan {{\theta }_{3}}+\tan {{\theta
}_{3}}\tan {{\theta }_{1}}=\dfrac{2a-h}{a}....\left( x \right)\]

which is similar to equation \[\left( vi \right)\]

Substitute \[\left( viii \right),\left( ix \right)\]and \[\left( x \right)\]in equation \[\left( vii
\right)\]

\[\dfrac{0-\left( \dfrac{-k}{a} \right)}{1-\left( \dfrac{2a-h}{a} \right)}=c\]

Simplifying the above equation,

\[\dfrac{\dfrac{k}{a}}{\dfrac{a-2a+h}{a}}=c\Rightarrow \dfrac{k}{h-a}=c\]

\[\therefore k=c\left( h-a \right)\]

\[k=hc-ac\]

Replace \[\left( h,k \right)\]by \[\left( x,y \right)\]

\[\therefore y=cx-ac\]where \[c\]is a constant

\[y=c\left( x-a \right)\]

This equation represents a straight line.

Note: We know that the problem is to prove the sum of the angles \[\left( {{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}} \right)\]made by them with the axis is constant. By taking \[\left( {{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}} \right)\]as constant, we can prove that the locus of points make a straight line.