Question

Let A(a,0) and B(b,0) be fixed distance points on the x-axis, none of which coincides with the origin O(0,0), and let C be a point on the y-axis. Let L be a line through the O(0,0) and perpendicular to the line AC. The locus of the point of intersection of the lines L and BC if C varies along the y-axis, is (provided $$c^2+ab\ne 0$$).
(a) $${\displaystyle \frac{x^2}{a}}+{\displaystyle \frac{y^2}{b}}=x$$
(b) $${\displaystyle \frac{x^2}{a}}+{\displaystyle \frac{y^2}{b}}=y$$
(c) $${\displaystyle \frac{x^2}{b}}+{\displaystyle \frac{y^2}{a}}=x$$
(d) $${\displaystyle \frac{x^2}{b}}+{\displaystyle \frac{y^2}{a}}=y$$

Answer 1

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 $$A(a,0)$$and $$B(b,0)$$ are two fixed points on x axis, let us assume the variable point C on y axis as $$(0,k)$$.

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

Then the slope of the line AC is given as:
$$m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$$
$$m={\displaystyle \frac{k-0}{0-a}}$$
$$m={\displaystyle \frac{-k}{a}}$$
Now the slope of the line perpendicular to line AC is $${\displaystyle \frac{-1}{m}}$$, 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:
$$y=\left({\displaystyle \frac{-1}{m}}\right)x$$
$$y=\left({\displaystyle \frac{a}{k}}\right)x$$
$$k={\displaystyle \frac{ax}{y}}$$
Now the equation of line BC can be found out using $${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=1$$(intercept form) where a and b are x-intercepts and y-intercept respectively.
Therefore, the equation of line BC is:
$${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{k}}=1$$
Substituting $$k={\displaystyle \frac{ax}{y}}$$ in the above equation we will have:
$${\displaystyle \frac{x}{b}}+{\displaystyle \frac{y}{\left({\displaystyle \frac{ax}{y}}\right)}}=1$$
$${\displaystyle \frac{x}{b}}+{\displaystyle \frac{y^2}{ax}}=1$$
$$\begin{array}{rl}& \\ & {\displaystyle \frac{x^2}{b}}+{\displaystyle \frac{y^2}{a}}=x\end{array}$$
Thus, the locus of point of intersection of L and BC is given as $${\displaystyle \frac{x^2}{b}}+{\displaystyle \frac{y^2}{a}}=x$$
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.