Question

The locus of the foot of perpendicular drawn from the centre of the ellipse ${{x}^{2}}+3{{y}^{2}}=6$ on any tangent to it is
(a) ${{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}=6{{x}^{2}}+2{{y}^{2}}$
(b) ${{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}=6{{x}^{2}}-2{{y}^{2}}$
(c) ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=6{{x}^{2}}+2{{y}^{2}}$
(d) ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=6{{x}^{2}}-2{{y}^{2}}$

Answer 1

First, before proceeding for this, by rearranging the given ellipse equation by dividing both sides by 6 to get standard form as $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ . Then, we must know the formula for the tangent of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is given by $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$ . Then, we need to find the equation of the line through the point (0, 0) and perpendicular to 1 is given by $y-0=\left( \dfrac{-1}{m} \right)x-0$ to get value of m and then the desired result.

Complete step-by-step answer:
In this question, we are supposed to find the locus of the foot of perpendicular drawn from the centre of the ellipse ${{x}^{2}}+3{{y}^{2}}=6$ on any tangent to it.

So, before proceeding for this, by rearranging the given ellipse equation by dividing both sides by 6, we get:
$\begin{aligned} & \dfrac{{{x}^{2}}}{6}+\dfrac{3{{y}^{2}}}{6}=\dfrac{6}{6} \\\ & \Rightarrow \dfrac{{{x}^{2}}}{6}+\dfrac{{{y}^{2}}}{2}=1 \\\ \end{aligned}$
Now, by comparing the above equation with the standard form of the ellipse as $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ , we get:
${{a}^{2}}=6,\text{ }{{b}^{2}}=2$
Then, we must know the formula for the tangent of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is given by:
$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
Then, by substituting the values we calculated above as ${{a}^{2}}=6,\text{ }{{b}^{2}}=2$ in the expression to get:
$y=mx+\sqrt{6{{m}^{2}}+2}$
Now, we need to find the equation of the line through the point (0, 0) and perpendicular to 1 is given by:
$\begin{aligned} & y-0=\left( \dfrac{-1}{m} \right)x-0 \\\ & \Rightarrow y=\dfrac{-x}{m} \\\ & \Rightarrow m=\dfrac{-x}{y} \\\ \end{aligned}$
Then, by substituting the value of m calculated above in the tangent equation, we get:
$y=\left( \dfrac{-x}{y} \right)x+\sqrt{6{{\left( \dfrac{-x}{y} \right)}^{2}}+2}$
Now, we need to solve the above equation until we get an expression similar to any of the options given as:
$\begin{aligned} & y=\left( \dfrac{-{{x}^{2}}}{y} \right)+\sqrt{6\left( \dfrac{{{x}^{2}}}{{{y}^{2}}} \right)+2} \\\ & \Rightarrow y=\left( \dfrac{-{{x}^{2}}}{y} \right)+\sqrt{\dfrac{6{{x}^{2}}+2{{y}^{2}}}{{{y}^{2}}}} \\\ & \Rightarrow y=\left( \dfrac{-{{x}^{2}}}{y} \right)+\dfrac{1}{y}\sqrt{6{{x}^{2}}+2{{y}^{2}}} \\\ & \Rightarrow {{y}^{2}}=-{{x}^{2}}+\sqrt{6{{x}^{2}}+2{{y}^{2}}} \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}=\sqrt{6{{x}^{2}}+2{{y}^{2}}} \\\ & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=6{{x}^{2}}+2{{y}^{2}} \\\ \end{aligned}$
So, the locus of the foot of perpendicular drawn from the centre of the ellipse ${{x}^{2}}+3{{y}^{2}}=6$ on any tangent to it is ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=6{{x}^{2}}+2{{y}^{2}}$ .

So, the correct answer is “Option (c)”.

Note: Now, to solve these type of the questions we need to know some of the basic fundamentals of the equation of the line given by $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where m is slope and $\left( {{x}_{1}},{{y}_{1}} \right)$ is any point where line passes. Also, when the line is perpendicular when the slope becomes the negative reciprocal of the given slope as if the slope is m then the slope of the perpendicular line is $\dfrac{-1}{m}$ .