Q.(a) Show that the sum of the focal distances of any point on an ellipse is equal to the length of the major axis and also prove that the locus of a point which moves so that the sum of its distances from $(3, 0)$ and $(-3, 0)$ is 9, is $\dfrac{x^2}{\left(\frac{81}{4}\right)} + \dfrac{y^2}{\left(\frac{45}{4}\right)} = 1$. OR

(a) proves the focal-distance sum property of an ellipse and uses the same idea to derive a locus equation; (b) maximises the area of a rectangle inscribed in a circle to show the maximum is $2r^2$.

(a) Focal distance sum, and the locus $PS+PS'=9$

1. Let the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b>0$), eccentricity $e$, foci $S(ae,0),S'(-ae,0)$, and $P(x,y)$ any point on it, so $y^2=b^2\Big(1-\dfrac{x^2}{a^2}\Big)$ and $b^2=a^2(1-e^2)$.
2. $SP^2=(x-ae)^2+y^2 = x^2-2aex+a^2e^2+b^2-\dfrac{b^2x^2}{a^2} = x^2\Big(1-\dfrac{b^2}{a^2}\Big)-2aex+a^2e^2+b^2 = x^2e^2-2aex+(a^2e^2+b^2)$.
3. Since $b^2=a^2-a^2e^2$, we get $a^2e^2+b^2=a^2$, so $SP^2=x^2e^2-2aex+a^2=(a-ex)^2$, hence $SP=a-ex$ (positive since $|ex|<a$ on the ellipse).
4. By the same computation with $S'(-ae,0)$: $S'P^2=(x+ae)^2+y^2=(a+ex)^2$, so $S'P=a+ex$.
5. Therefore $SP+S'P=(a-ex)+(a+ex)=2a$ = length of the major axis. Hence proved.
6. Locus: Let $P(x,y)$, $S(3,0)$, $S'(-3,0)$, with $PS+PS'=9$. Then $PS=\sqrt{(x-3)^2+y^2}$, $PS'=\sqrt{(x+3)^2+y^2}$, and $PS=9-PS'$.
7. Squaring: $(x-3)^2+y^2=81-18\,PS'+(x+3)^2+y^2$. The $y^2$ cancels; $(x-3)^2-(x+3)^2=-12x$, so $-12x=81-18\,PS'\Rightarrow PS'=\dfrac{81+12x}{18}=\dfrac92+\dfrac23x$.
8. Square again: $(x+3)^2+y^2=\Big(\dfrac92+\dfrac23x\Big)^2=\dfrac{81}{4}+6x+\dfrac49x^2$.
9. Expand the left side and simplify: $x^2+6x+9+y^2=\dfrac{81}{4}+6x+\dfrac49x^2 \Rightarrow x^2\Big(1-\dfrac49\Big)+y^2=\dfrac{81}{4}-9 \Rightarrow \dfrac59x^2+y^2=\dfrac{45}{4}$.
10. Divide by $\dfrac{45}{4}$: $\dfrac{x^2}{(45/4)(9/5)}+\dfrac{y^2}{45/4}=1$, and $\dfrac{45}{4}\cdot\dfrac95=\dfrac{81}{4}$, giving $\dfrac{x^2}{81/4}+\dfrac{y^2}{45/4}=1$, exactly as required.

(b) Largest rectangle inscribed in a circle of radius $r$

1. Centre the circle at the origin; inscribe a rectangle with sides $2x$ and $2y$ (parallel to the axes), so its vertices $(\pm x,\pm y)$ lie on the circle: $x^2+y^2=r^2$.
2. Area $A=(2x)(2y)=4xy$. Using $y=\sqrt{r^2-x^2}$: $A(x)=4x\sqrt{r^2-x^2}$, $0<x<r$. …