Question

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ from any of its foci?

• A. $\left(-1,\sqrt{3}\right)$
• B. $\left(-2,\sqrt{3}\right)$
• C. $\left(-1,\sqrt{2}\right)$
• D. $\left(1,2\right)$

Answer 1

The correct option is A

$\left(-1,\sqrt{3}\right)$

Explanation for the correct option:

Step 1: Compare $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ with the given equation.

$\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{2}=1$

$$\begin{gathered} {\mathrm{a}}^2=4 \\ \Rightarrow \mathrm{a}=2 \end{gathered}$$

$$\begin{gathered} {\mathrm{b}}^2=2 \\ \Rightarrow \mathrm{b}=\sqrt{2} \end{gathered}$$

Step 2: Find the eccentricity.

The formula for the eccentricity of an ellipse is

$\mathrm{e}=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}$

$\Rightarrow \mathrm{e}=\sqrt{1-\frac{2}{4}}$

$\Rightarrow \mathrm{e}=\frac{1}{\sqrt{2}}$

Step 3: Find the focus.

Focus is given as $\left(\mathrm{ae},0\right)$.

Focus $=\left(\frac{2}{\sqrt{2}},0\right)=\left(\sqrt{2},0\right)$ [Rationalizing the denominator]

Step 4: Derive the equation of the tangent.

The equation of the tangent is given as

$\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$

$\Rightarrow \mathrm{y}=\mathrm{mx}+\sqrt{4{\mathrm{m}}^2+2}$

The line passes through $\left(\mathrm{h},\mathrm{k}\right)$.

$\Rightarrow {\left(\mathrm{k}-\mathrm{mh}\right)}^2=4{\mathrm{m}}^2+2$ $...\left(\mathrm{i}\right)$

Step 5: Derive the equation of the perpendicular.

The slope of the perpendicular is given as $-\frac{1}{\mathrm{m}}$.

The equation of the perpendicular line is given as

$\mathrm{y}=\left(-\frac{1}{\mathrm{m}}\right)\left(\mathrm{x}-\sqrt{2}\right)$

$\Rightarrow -\mathrm{my}=\left(\mathrm{x}-\sqrt{2}\right)$

$\Rightarrow \mathrm{x}+\mathrm{my}=\sqrt{2}$

$\Rightarrow {\left(\mathrm{x}+\mathrm{my}\right)}^2=2$

$\Rightarrow {\left(\mathrm{h}+\mathrm{mk}\right)}^2=2$ $...\left(\mathrm{ii}\right)$

Step 6: Add both the equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ to find the locus.

${\left(\mathrm{k}-\mathrm{mh}\right)}^2+{\left(\mathrm{h}+\mathrm{mk}\right)}^2=4{\mathrm{m}}^2+2+2$

$\Rightarrow {\mathrm{k}}^2+{\mathrm{m}}^2{\mathrm{h}}^2-2\mathrm{mhk}+{\mathrm{h}}^2+{\mathrm{m}}^2{\mathrm{k}}^2+2\mathrm{mhk}=4{\mathrm{m}}^2+4$

$\Rightarrow {\mathrm{h}}^2\left(1+{\mathrm{m}}^2\right)+{\mathrm{k}}^2\left(1+{\mathrm{m}}^2\right)=4\left(1+{\mathrm{m}}^2\right)$

$\Rightarrow \left(1+{\mathrm{m}}^2\right)\left({\mathrm{h}}^2+{\mathrm{k}}^2\right)=4\left(1+{\mathrm{m}}^2\right)$

$\Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2=4$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=4$ [Equation of a circle]

Step 7: Put the coordinates in the equation.

${\mathrm{x}}^2+{\mathrm{y}}^2=4$

$\Rightarrow {\left(-1\right)}^2+{\left(\sqrt{3}\right)}^2=4$

$\Rightarrow 4=4$

Since the left-hand side is equal to the right-hand side, $\left(-1,\sqrt{3}\right)$ lies on the locus of the foot of the perpendicular.

The correct answer is option a) $\left(-1,\sqrt{3}\right)$.