Question

Find the locus of a point \[O\]when the three normals drawn from it are such that two of them make equal angles with the given line \[y=mx+c\].

Answer 1

Hint: Take the directrix as \[x=-a\]and find the normal of the tangent by taking the slope as \['m'\].

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 three normals be drawn whose 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.

Let the equation of parabola be \[\Rightarrow {{y}^{2}}=4ax....\left( i \right)\]

Any normal to the equation \[\left( i \right)\]can be put as

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

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

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

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

The equation \[\left( iv \right)\]is obtained from the equation of \[3\] normals to any point to the parabola The product of the roots will be \[\left( \dfrac{-k}{a} \right)\].

Consider equation \[\left( ii \right)\], \[y=mx-2am-a{{m}^{3}}\]

Put \[\left( x,y \right)\]as \[\left( h,k \right)\].

\[\therefore \]We can modify the equations as

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

Rearranging the equation, we get

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

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

Substitute the value \[m\]from equation \[\left( iv \right)\]to equation \[\left( v \right)\]

\[{{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}\Rightarrow m=\dfrac{k}{a}\]

\[a{{\left( \dfrac{k}{a} \right)}^{3}}+\dfrac{k}{a}\left( 2a-h \right)+k=0\]

\[\dfrac{{{k}^{3}}}{{{a}^{2}}}+\dfrac{k}{a}\left( 2a-h \right)+k=0\]

Canceling out\[k\], by dividing with\[k\],

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

Multiply throughout by \[{{a}^{2}}\]

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

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

\[\therefore {{k}^{2}}+2{{a}^{2}}-ah+{{a}^{2}}=0\]

\[{{k}^{2}}+3{{a}^{2}}-ah=0\]

\[{{k}^{2}}=ah-3{{a}^{2}}\]

\[\Rightarrow {{k}^{2}}=a\left( h-3a \right)\]

Put \[\left( x,y \right)=\left( h,k \right)\]

\[\Rightarrow {{y}^{2}}=a\left( h-3a \right)\]

\[\therefore \]Locus of \[\left( h,k \right)\]is \[{{y}^{2}}=a\left( x-3a \right)\].

Note: Remember that the tangents at the extremities of a focal chord intersect at the right angle to the directrix. The equation of tangents connecting the three normals play an important role in solving the question i.e. equation \[\left( ii \right),\left( iii \right)\] and \[\left( iv \right)\].