Answer

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Hint: Assume a variable point C on y axis as (0,k) and find the slope of the line AC. Using the perpendicularity condition of lines, find the slope of the line C and thus the equation of line L. Finally, substitute the expression of \[k\] that you have obtained from L on the line BC for the required locus.

Given \[A\left( a,0 \right)\]and \[B\left( b,0 \right)\] are two fixed points on x axis, let us assume the variable point C on y axis as \[\left( 0,k \right)\].

Plotting the diagram with the above data, we will have it as:

Then the slope of the line AC is given as:

\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]

\[m=\dfrac{k-0}{0-a}\]

\[m=\dfrac{-k}{a}\]

Now the slope of the line perpendicular to line AC is \[\dfrac{-1}{m}\], since the product of slopes of perpendicular lines is -1.

Therefore, the equation of line L passing through origin and perpendicular to AC is given as:

\[y=\left( \dfrac{-1}{m} \right)x\]

\[y=\left( \dfrac{a}{k} \right)x\]

\[k=\dfrac{ax}{y}\]

Now the equation of line BC can be found out using \[\dfrac{x}{a}+\dfrac{y}{b}=1\](intercept form) where a and b are x-intercepts and y-intercept respectively.

Therefore, the equation of line BC is:

\[\dfrac{x}{a}+\dfrac{y}{k}=1\]

Substituting \[k=\dfrac{ax}{y}\] in the above equation we will have:

\[\dfrac{x}{b}+\dfrac{y}{\left( \dfrac{ax}{y} \right)}=1\]

\[\dfrac{x}{b}+\dfrac{{{y}^{2}}}{ax}=1\]

\[\begin{align}

& \\

& \dfrac{{{x}^{2}}}{b}+\dfrac{{{y}^{2}}}{a}=x \\

\end{align}\]

Thus, the locus of point of intersection of L and BC is given as \[\dfrac{{{x}^{2}}}{b}+\dfrac{{{y}^{2}}}{a}=x\]

Hence, option A is the correct answer.

Note: For any given two lines having slopes \[{{m}_{1}}\] and \[{{m}_{2}}\], then the condition for them to be parallel is \[{{m}_{1}}={{m}_{2}}\] and the condition to be perpendicular is \[{{m}_{1}}.{{m}_{2}}=-1\]. Also, \[\frac{x}{a}+\frac{y}{b}=1\], is the intercept form of a line where a and b are x-intercept and y-intercept respectively.

Given \[A\left( a,0 \right)\]and \[B\left( b,0 \right)\] are two fixed points on x axis, let us assume the variable point C on y axis as \[\left( 0,k \right)\].

Plotting the diagram with the above data, we will have it as:

Then the slope of the line AC is given as:

\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]

\[m=\dfrac{k-0}{0-a}\]

\[m=\dfrac{-k}{a}\]

Now the slope of the line perpendicular to line AC is \[\dfrac{-1}{m}\], since the product of slopes of perpendicular lines is -1.

Therefore, the equation of line L passing through origin and perpendicular to AC is given as:

\[y=\left( \dfrac{-1}{m} \right)x\]

\[y=\left( \dfrac{a}{k} \right)x\]

\[k=\dfrac{ax}{y}\]

Now the equation of line BC can be found out using \[\dfrac{x}{a}+\dfrac{y}{b}=1\](intercept form) where a and b are x-intercepts and y-intercept respectively.

Therefore, the equation of line BC is:

\[\dfrac{x}{a}+\dfrac{y}{k}=1\]

Substituting \[k=\dfrac{ax}{y}\] in the above equation we will have:

\[\dfrac{x}{b}+\dfrac{y}{\left( \dfrac{ax}{y} \right)}=1\]

\[\dfrac{x}{b}+\dfrac{{{y}^{2}}}{ax}=1\]

\[\begin{align}

& \\

& \dfrac{{{x}^{2}}}{b}+\dfrac{{{y}^{2}}}{a}=x \\

\end{align}\]

Thus, the locus of point of intersection of L and BC is given as \[\dfrac{{{x}^{2}}}{b}+\dfrac{{{y}^{2}}}{a}=x\]

Hence, option A is the correct answer.

Note: For any given two lines having slopes \[{{m}_{1}}\] and \[{{m}_{2}}\], then the condition for them to be parallel is \[{{m}_{1}}={{m}_{2}}\] and the condition to be perpendicular is \[{{m}_{1}}.{{m}_{2}}=-1\]. Also, \[\frac{x}{a}+\frac{y}{b}=1\], is the intercept form of a line where a and b are x-intercept and y-intercept respectively.

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