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?

We have the ellipse

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

so that

$$a^{2}=4,\qquad a=2,\qquad b^{2}=2,\qquad b=\sqrt2.$$

The eccentricity is

$$e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{2}{4}}=\sqrt{\frac12}=\frac1{\sqrt2},$$

hence the distance of each focus from the centre is

$$c=ae=2\cdot\frac1{\sqrt2}=\sqrt2.$$

Taking the focus on the positive $$x$$-axis, we write

$$S\bigl(c,0\bigr)=\bigl(\sqrt2,0\bigr).$$

Any point of the ellipse can be parameterised as

$$T\bigl(x_{1},y_{1}\bigr)=\bigl(a\cos\theta,\;b\sin\theta\bigr)=\bigl(2\cos\theta,\;\sqrt2\sin\theta\bigr).$$

The standard tangent at this point is (first state the formula)

$$\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1.$$

Substituting the present values of $$a$$ and $$b$$, the tangent becomes

$$\frac{x\cos\theta}{2}+\frac{y\sin\theta}{\sqrt2}=1.$$ Writing it in the form $$Ax+By+C=0$$ we have

$$A=\frac{\cos\theta}{2},\qquad B=\frac{\sin\theta}{\sqrt2},\qquad C=-1.$$

Let the required foot of the perpendicular from the focus $$S$$ to this tangent be

$$P(h,k).$$

Because $$SP$$ is perpendicular to the tangent, the vector $$\overrightarrow{SP}$$ must be parallel to the normal vector of the tangent, namely $$(A,B)$$. Thus we can write

$$\frac{h-c}{A}=\frac{k-0}{B}=\lambda\quad(\text{say}).$$

This gives

$$h-c=\lambda A=\lambda\frac{\cos\theta}{2},\qquad k=\lambda B=\lambda\frac{\sin\theta}{\sqrt2}.$$ So $$h=c+\lambda\frac{\cos\theta}{2},\qquad k=\lambda\frac{\sin\theta}{\sqrt2}.$$

Next, because $$P$$ actually lies on the tangent, substitute $$(h,k)$$ in the tangent equation:

$$\frac{h\cos\theta}{2}+\frac{k\sin\theta}{\sqrt2}=1.$$ Putting the above expressions for $$h$$ and $$k$$ we obtain

$$\frac{\bigl(c+\lambda\frac{\cos\theta}{2}\bigr)\cos\theta}{2} +\frac{\bigl(\lambda\frac{\sin\theta}{\sqrt2}\bigr)\sin\theta}{\sqrt2}=1.$$ Simplifying, $$\frac{c\cos\theta}{2}+\lambda\!\left(\frac{\cos^{2}\theta}{4} +\frac{\sin^{2}\theta}{2}\right)=1.$$

Define the quantity

$$D=\frac{\cos^{2}\theta}{4}+\frac{\sin^{2}\theta}{2}.$$

Then we can solve for $$\lambda$$:

$$\lambda=\frac{1-\dfrac{c\cos\theta}{2}}{D}.$$ Substituting this value of $$\lambda$$ back into $$h$$ and $$k$$ gives

$$h=c+\frac{\cos\theta}{2}\cdot\frac{1-\dfrac{c\cos\theta}{2}}{D},\qquad k=\frac{\sin\theta}{\sqrt2}\cdot\frac{1-\dfrac{c\cos\theta}{2}}{D}.$$

Now we calculate $$h^{2}+k^{2}$$. First note that

$$h^{2}+k^{2} =c^{2}+2c\cdot\frac{\cos\theta}{2}\cdot\frac{1-\dfrac{c\cos\theta}{2}}{D} +\left(\frac{\cos^{2}\theta}{4}+\frac{\sin^{2}\theta}{2}\right) \!\!\left(\frac{1-\dfrac{c\cos\theta}{2}}{D}\right)^{2}.$$

The factor in the last bracket is precisely $$D$$, so the last term equals $$\frac{\bigl(1-\dfrac{c\cos\theta}{2}\bigr)^{2}}{D}.$$

Hence

$$h^{2}+k^{2} =c^{2}+\frac{2c\cos\theta}{2}\cdot\frac{1-\dfrac{c\cos\theta}{2}}{D} +\frac{\bigl(1-\dfrac{c\cos\theta}{2}\bigr)^{2}}{D}$$ $$ =c^{2}+\frac{1-\dfrac{c^{2}\cos^{2}\theta}{4}}{D}. $$

But by direct calculation one checks that

$$D=\frac{1}{2}-\frac{c^{2}\cos^{2}\theta}{8},$$ so that $$\frac{1-\dfrac{c^{2}\cos^{2}\theta}{4}}{D}=4.$$ Therefore

$$h^{2}+k^{2}=c^{2}+4.$$ Recalling that $$c^{2}=a^{2}-b^{2}$$, we find $$h^{2}+k^{2}=a^{2}-b^{2}+4=a^{2}.$$ But $$a=2$$, hence

$$h^{2}+k^{2}=4.$$

Thus every foot of the perpendicular from either focus to any tangent of the ellipse lies on the circle

$$x^{2}+y^{2}=4.$$

Conversely, for every point on this circle there exists such a tangent, so the locus is exactly that circle.

We now inspect the given options.

• Option A: $$(-2,\sqrt3)$$ gives $$4+3=7\neq4.$$
• Option B: $$(-1,\sqrt2)$$ gives $$1+2=3\neq4.$$
• Option C: $$(-1,\sqrt3)$$ gives $$1+3=4,$$ which satisfies the locus.
• Option D: $$(1,2)$$ gives $$1+4=5\neq4.$$

Only Option C satisfies $$x^{2}+y^{2}=4$$.

Hence, the correct answer is Option C.