Q.Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Concept and Intuition

For an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$): its own area is $\pi ab$; its auxiliary circle (drawn on the major axis as diameter) has radius $a$, area $\pi a^2$; its director circle (locus of points from which perpendicular tangents can be drawn) has radius $\sqrt{a^2+b^2}$, area $\pi(a^2+b^2)$.

Step-by-Step Solution

1. $x^2+4y^2-4=0 \implies x^2+4y^2=4 \implies \dfrac{x^2}{4}+\dfrac{y^2}{1}=1$. So $a^2=4\ (a=2)$, $b^2=1\ (b=1)$.
2. $A_1$ (ellipse area) $=\pi ab = \pi\times2\times1=2\pi$.
3. $A_2$ (director circle area) $=\pi(a^2+b^2)=\pi(4+1)=5\pi$.
4. $A_3$ (auxiliary circle area) $=\pi a^2 = 4\pi$.
5. $A_2+A_3-A_1 = 5\pi+4\pi-2\pi = 7\pi.$

Common Mistakes

• Swapping which of $a,b$ is the semi-major axis (must have $a^2=4>b^2=1$, so $a=2$).
• Using $\pi(a^2-b^2)$ or another wrong combination for the director circle's radius formula.

ANSWER: C

Concept and Intuition

The tangent to an ellipse at $(x_1,y_1)$ is $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$. Because the four latus-rectum end-points are symmetric about both axes, the four tangent lines automatically form a rhombus whose diagonals lie along the coordinate axes — so we only need each tangent's axis intercepts.

Step-by-Step Solution

1. Here $a^2=9,\,b^2=5\Rightarrow a=3,\,b=\sqrt5$, and $c^2=a^2-b^2=4\Rightarrow c=2$.
2. Latus-rectum endpoints: $(\pm c,\pm b^2/a)=(\pm2,\pm5/3)$.
3. Tangent at $(2,5/3)$: $\dfrac{2x}{9}+\dfrac{(5/3)y}{5}=1\Rightarrow\dfrac{2x}{9}+\dfrac{y}{3}=1$.
4. Its $x$-intercept ($y=0$): $x=9/2$; its $y$-intercept ($x=0$): $y=3$.
5. By the symmetry of the four points $(\pm2,\pm5/3)$, intersecting adjacent tangent lines (checked directly) gives quadrilateral vertices exactly at $(9/2,0),(0,3),(-9/2,0),(0,-3)$ — a rhombus with diagonals $9$ (horizontal) and $6$ (vertical).
6. Area of a rhombus $=\dfrac12 d_1 d_2=\dfrac12(9)(6)=27$.

Common Mistakes

• Forgetting the factor of $\tfrac12$ in the rhombus-area formula.
• Mixing up which axis intercept comes from which coefficient.

ANSWER: A

Concept and Intuition

Taking the midpoint of a fixed point $F$ and a point $P$ tracing a curve is a scaling (homothety) of that curve by factor $\tfrac12$ centred at $F$. A linear scaling by factor $k$ shrinks every length by $k$ and every area by $k^2$ — so the midpoint locus here is an ellipse similar to the original but with half the semi-axes, hence a quarter of the area, regardless of where the focus actually sits.

Step-by-Step Solution

1. Area of the given ellipse: $A_1=\pi a b$.
2. Let $F=(c,0)$ be a focus (fixed) and $P=(x,y)$ a variable point on the ellipse. The midpoint is $M=\left(\dfrac{x+c}2,\ \dfrac{y}2\right)$.
3. As $P$ traces $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, writing $X=\dfrac{x+c}2,\ Y=\dfrac{y}2$ gives $x=2X-c,\ y=2Y$, substituting: $\dfrac{(2X-c)^2}{a^2}+\dfrac{(2Y)^2}{b^2}=1$, which is an ellipse in $X,Y$ with semi-axes $\dfrac{a}2$ and $\dfrac{b}2$ (centred at $\dfrac{c}2$ on the X-axis).
4. Its area: $A_2=\pi\cdot\dfrac{a}2\cdot\dfrac{b}2=\dfrac{\pi ab}4$.
5. $A_1:A_2 = \pi ab : \dfrac{\pi ab}4 = 4:1$.

Common Mistakes

• Forgetting the midpoint transformation scales area by the square of the linear scale factor $\left(\tfrac12\right)^2=\tfrac14$, not by $\tfrac12$ itself.
• Getting distracted by the shift due to the fixed focus $c$, which affects only the position, not the size, of the locus ellipse.

ANSWER: C

Concept and Intuition

For ellipse $x^2/a^2+y^2/b^2=1$ with $a>b$, foci are at $(\pm ae,0)$ and the endpoints of the minor axis are $(0,\pm b)$. The triangle formed by the two foci and one minor-axis endpoint has the segment joining the foci (length $2ae$) as base and $b$ as height (perpendicular distance from $(0,b)$ to the $x$-axis).

Step-by-Step Solution

1. Area $=\tfrac12\times2ae\times b=aeb=\sqrt3$. With $a=2$: $2eb=\sqrt3\Rightarrow eb=\sqrt3/2$.
2. Also $b^2=a^2(1-e^2)=4(1-e^2)\Rightarrow b=2\sqrt{1-e^2}$.
3. Substitute: $e\cdot2\sqrt{1-e^2}=\sqrt3/2\Rightarrow e\sqrt{1-e^2}=\sqrt3/4$.
4. Square: $e^2(1-e^2)=3/16$. Let $t=e^2$: $t-t^2=3/16\Rightarrow16t^2-16t+3=0$.
5. $t=\dfrac{16\pm\sqrt{256-192}}{32}=\dfrac{16\pm8}{32}=3/4$ or $1/4$.
6. So $e=\sqrt3/2$ or $e=1/2$ (both valid since $b<2$ holds in each case).

Common Mistakes

• Using $c$ (focal distance) instead of $ae$ directly, or forgetting $a=2$ specifically here.
• Discarding one of the two valid roots of the quadratic in $e^2$ — both satisfy $b<a=2$.

ANSWER: A

Concept and Intuition

Joining $(\pm a,0)$ and $(0,\pm b)$ produces a rhombus whose diagonals are the major axis

($2a$) and minor axis ($2b$); a rhombus's area is half the product of its diagonals, i.e.

$\tfrac12(2a)(2b)=2ab$. The foci are at $(\pm ae,0)$, so the distance between them is $2ae$.

These two given numbers pin down $a$ and $b$ (hence $e$) completely.

Step-by-Step Solution

1. Area of quadrilateral $=2ab=8\sqrt3 \Rightarrow ab=4\sqrt3$.
2. Distance between foci $=2ae=4\sqrt2 \Rightarrow ae=2\sqrt2$.
3. For an ellipse, $b^2=a^2(1-e^2)=a^2-(ae)^2=a^2-8$.
4. Also $b=\dfrac{4\sqrt3}{a}\Rightarrow b^2=\dfrac{48}{a^2}$.
5. Equate: $\dfrac{48}{a^2}=a^2-8 \Rightarrow a^4-8a^2-48=0$.
6. Let $u=a^2$: $u^2-8u-48=0 \Rightarrow u=\dfrac{8\pm\sqrt{64+192}}{2}=\dfrac{8\pm16}{2}$, so $u=12$ (rejecting the negative root).
7. $a^2=12 \Rightarrow b^2=12-8=4$.
8. $e^2=1-\dfrac{b^2}{a^2}=1-\dfrac{4}{12}=\dfrac{2}{3} \Rightarrow e=\sqrt{\dfrac{2}{3}}$.

Common Mistakes

• Using area $=4ab$ or $ab$ instead of the correct rhombus formula $2ab$.
• Forgetting $b^2=a^2-(ae)^2$ requires foci distance divided by 2 to get $ae$ (not $2ae$ itself).

ANSWER: D