Question

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

Answer 1

Hint: First, before proceeding for this, by rearranging the given ellipse equation by dividing both sides by 6 to get standard form as ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ . Then, we must know the formula for the tangent of the ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ 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 $y-0=\left({\displaystyle \frac{-1}{m}}\right)x-0$ 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 $x^2+3y^2=6$ on any tangent to it.

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

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 $y-y_1=m(x-x_1)$ where m is slope and $(x_1,y_1)$ 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 ${\displaystyle \frac{-1}{m}}$ .