Q.The area of the region bounded by the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is
(A) $20\pi$ sq units
(B) $20\pi^2$ sq units
(C) $16\pi^2$ sq units
(D) $25\pi$ sq units

Concept & Intuition

The latus rectum of an ellipse is a chord through a focus perpendicular to the major axis. Its endpoints are symmetric about both axes. Tangents at these four points will be symmetric as well, so the quadrilateral they form is actually a rhombus centered at the origin, with its diagonals along the coordinate axes. The area of such a rhombus is simply half the product of the lengths of its diagonals — which here are the intercepts cut off on the x‑axis and y‑axis by the four tangents. So we just need the equations of the tangents at one latus‑rectum endpoint, find where they hit the axes, and double appropriately.

Step‑by‑step solution

1. Identify the ellipse parameters

   The ellipse is $\frac{x^2}{9} + \frac{y^2}{5} = 1$.

   So $a^2 = 9 \Rightarrow a = 3$, $b^2 = 5 \Rightarrow b = \sqrt{5}$.

   Eccentricity: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

   Foci are at $(\pm ae, 0) = (\pm 2, 0)$.

2. Endpoints of the latus recta

   For the right focus $(2,0)$, the latus rectum is the vertical line $x = 2$.

   Substitute into the ellipse: $\frac{4}{9} + \frac{y^2}{5} = 1 \Rightarrow \frac{y^2}{5} = \frac{5}{9} \Rightarrow y^2 = \frac{25}{9} \Rightarrow y = \pm \frac{5}{3}$.

   So the four endpoints are:

3. Equation of the tangent at one endpoint For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the tangent at $(x_1, y_1)$ is

Take $(x_1, y_1) = (2, \frac{5}{3})$. Then

Multiply by 9: $2x + 3y = 9$.

So the tangent line is $2x + 3y = 9$.

4. Find the intercepts of this tangent

   • x‑intercept (set $y=0$): $2x = 9 \Rightarrow x = \frac{9}{2}$.
   • y‑intercept (set $x=0$): $3y = 9 \Rightarrow y = 3$.

   By symmetry, the tangent at $(2, -\frac{5}{3})$ will be $2x - 3y = 9$, giving x‑intercept $\frac{9}{2}$ and y‑intercept $-3$.

   The tangents at the left‑side endpoints $(-2, \pm \frac{5}{3})$ will be $-2x \pm 3y = 9$, giving x‑intercept $-\frac{9}{2}$ and y‑intercepts $\pm 3$.

5. Shape of the quadrilateral

   The four tangents are:

   $\begin{aligned} &2x + 3y = 9, \quad 2x - 3y = 9, \\ &-2x + 3y = 9, \quad -2x - 3y = 9. \end{aligned}$ …

Step 1 — Foci of the first ellipse. $\dfrac{x^2}{9}+\dfrac{y^2}{5}=1$ has $a^2=9,\ b^2=5$, major axis along $x$, so $c^2=a^2-b^2=4,\ c=2$. Foci: $(\pm 2, 0)$.

Step 2 — Foci of the second ellipse. $\dfrac{x^2}{5}+\dfrac{y^2}{9}=1$ has larger denominator under $y^2$, so major axis along $y$, $c^2=9-5=4,\ c=2$. Foci: $(0, \pm 2)$. …

The key idea: a circle touching an ellipse internally means the circle lies completely inside the ellipse and meets it at exactly one point. For a circle centred at $(1,0)$, the ellipse is centred at $(0,0)$ with semi-axes $4$ (along $x$) and $2$ (along $y$). Since the circle’s centre is on the $x$-axis, the point of tangency will also lie on the $x$-axis — the line joining the centres is the natural direction for the shortest distance from the centre to the ellipse.

1. Write the ellipse in standard form.

   $x^2 + 4y^2 = 16$ becomes $\frac{x^2}{16} + \frac{y^2}{4} = 1$. So $a = 4$, $b = 2$.

2. Identify the geometry of internal tangency.

   The circle $(x-1)^2 + y^2 = r^2$ has centre $C = (1,0)$. For internal tangency, the circle must be inside the ellipse, so the distance from $C$ to the ellipse along the $x$-axis (the line through both centres) must equal $r$. The ellipse extends from $x = -4$ to $x = 4$ on the $x$-axis. Since $C$ is at $x=1$, the nearest point on the ellipse to the right is at $x=4$ — but that’s too far. Actually, the point of tangency will be somewhere between $C$ and the ellipse’s boundary on the same side.

   Because the circle’s centre is inside the ellipse (check: $1^2 + 0 < 16$), the circle can expand until it just touches the ellipse from inside. The tangency point lies on the line joining the centres — here the $x$-axis — so let the point be $(x_0, 0)$ on the ellipse. Then $x_0^2 = 16$, so $x_0 = 4$ or $x_0 = -4$. But $C$ is at $x=1$, so the nearer boundary is $x=4$. That would give $r = 4 - 1 = 3$, which is not among the options. So the tangency is not at the vertex — it occurs at some other point where the circle’s curvature matches the ellipse’s curvature? No, that’s for osculating circles. For simple tangency, the condition is that the circle and ellipse share a common tangent at the point of contact.

3. Set up the tangency condition.

   Let the point of contact be $(x_1, y_1)$ on the ellipse. Then:

   • It lies on the ellipse: $x_1^2 + 4y_1^2 = 16$.
   • It lies on the circle: $(x_1 - 1)^2 + y_1^2 = r^2$.
   • The gradients (slopes of tangents) are equal at that point.

   For the ellipse $x^2 + 4y^2 = 16$, differentiate implicitly: $2x + 8y y' = 0 \Rightarrow y' = -\frac{x}{4y}$.

   For the circle $(x-1)^2 + y^2 = r^2$, differentiate: $2(x-1) + 2y y' = 0 \Rightarrow y' = -\frac{x-1}{y}$.

   Equate slopes at $(x_1, y_1)$:

Assuming $y_1 \neq 0$, multiply both sides by $-y_1$:

Multiply by 4: $x_1 = 4x_1 - 4 \Rightarrow 3x_1 = 4 \Rightarrow x_1 = \frac{4}{3}$.

4. Find $y_1$ from the ellipse. …

We start with the given equation:

This is a quadratic in $x$ and $y$ with both squared terms positive, so it’s an ellipse. The key is to rewrite it in standard form by completing the square — that reveals the center, orientation, and lengths of axes, which then give the directrices.

1. Complete the square for $x$ and $y$.

Group the $x$ terms and $y$ terms:

Factor out the coefficients of the squares:

Complete each square:

• For $x^2 - 2x$: add and subtract $1$ → $(x-1)^2 - 1$
• For $y^2 - 4y$: add and subtract $4$ → $(y-2)^2 - 4$

Substitute:

Divide through by 36:

2. Identify the orientation and parameters.

The standard form is $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$ with $a > b$ for a vertical major axis.

Here:

• Center: $(h,k) = (1,2)$
• $a^2 = 9$ → $a = 3$ (semi-major axis, along $y$-direction)
• $b^2 = 4$ → $b = 2$ (semi-minor axis, along $x$-direction)

Since the larger denominator is under the $y$-term, the major axis is vertical.

3. Find the eccentricity $e$.

For an ellipse, $e = \sqrt{1 - \frac{b^2}{a^2}}$:

4. Determine the directrices.

For a vertical ellipse, the directrices are horizontal lines given by:

The key here is to recall the standard definitions and parameters of a hyperbola. Every hyperbola has a transverse axis (length $2a$), a conjugate axis (length $2b$), foci at distance $c$ from the centre, and directrices at distance $a/e$ from the centre. The eccentricity $e$ is defined by $c = ae$, and the relation $c^2 = a^2 + b^2$ always holds. The distance between the two foci is $2c$, and the distance between the two directrices is $2a/e$.

Once you translate each condition into an equation involving $a$, $b$, $c$, and $e$, the eccentricities come out cleanly.

1. First hyperbola: The distance between its foci is $2c$. The distance between its directrices is $2a/e_1$. The condition says $2c = 2 \times (2a/e_1)$, i.e. $c = 2a/e_1$. But we also know $c = a e_1$. Equate:

Cancel $a$ (non-zero) and multiply: $e_1^2 = 2$, so $e_1 = \sqrt{2}$ (eccentricity is positive). …

The concept first

For $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $b<a$, the foci sit at $(\pm ae,0)$ and the corresponding directrices at $x=\pm\dfrac{a}{e}$ (the directrix on the same side as its focus). So the perpendicular distance from a focus to its own directrix is

That single relation contains everything the question hides: it pins down $e$, and $e$ pins down $b$.

Step-by-step

Step 1 — set up the focus–directrix equation. Here $a^2=9\Rightarrow a=3$:

Multiply through by $\sqrt5\,e$:

Step 2 — solve for $e$.

Eccentricity is positive, so $e=\dfrac{10}{6\sqrt5}=\dfrac{5}{3\sqrt5}=\dfrac{\sqrt5}{3}$.

Step 3 — get $b$.

and indeed $b=2<3$, the condition stated in the question. The ellipse is

Step 4 — check the point lies on it. The point is $\left(\dfrac{3}{\sqrt2},\dfrac{b}{\sqrt2}\right)=\left(\dfrac{3}{\sqrt2},\dfrac{2}{\sqrt2}\right)$: …

First, we need to identify the conic. The equation is

Expand the right-hand side:

Bring all terms to one side:

Factor $y^2$:

Divide through by $a^2 (a^2 + b^2)$:

This is an ellipse centred at the origin, with semi-major axis along the $x$-axis of length $\sqrt{a^2+b^2}$ and semi-minor axis along the $y$-axis of length $a$.

Now, for an ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ with $A > B$, the eccentricity is

The foci are at $(\pm A e, 0) = \left(\pm \sqrt{a^2+b^2} \cdot \frac{b}{\sqrt{a^2+b^2}}, 0\right) = (\pm b, 0)$.

The directrices are the lines $x = \pm \frac{A}{e} = \pm \frac{\sqrt{a^2+b^2}}{b/\sqrt{a^2+b^2}} = \pm \frac{a^2+b^2}{b}$.

The focus $S$ is one of $(\pm b, 0)$; the nearer directrix to $S$ is the one on the same side of the centre. For $S = (b, 0)$, the nearer directrix is $x = \frac{a^2+b^2}{b}$. …