Let e₁ and e₂ be the eccentricities of the ellipse x²/b² + y²/25 = 1 and the hyperbola x²/16 - y²/b² = 1, respectively. If b is less than 5 and e₁e₂ = 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

 

❓ Question

Let $e_{1}$ and $e_{2}$ be the eccentricities of

$$\text{Ellipse: }\frac{x^2}{b^2}+\frac{y^2}{25}=1$$

and

$$\text{Hyperbola: }\frac{x^2}{16}-\frac{y^2}{b^2}=1$$

respectively.
If $b<5$ and $e_{1}e_{2}=1$, then find the eccentricity of the ellipse whose axes are along the coordinate axes and which passes through all four foci (two of the given ellipse and two of the hyperbola).

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Find $e_1$ and $e_2$ relations

For ellipse: $\dfrac{x^2}{b^2} + \dfrac{y^2}{25} = 1$

Major axis = along $y$-axis (since $25 > b^2$).
Eccentricity:

$$e_1 = \sqrt{1 - \frac{b^2}{25}}.$$

For hyperbola: ${\displaystyle \frac{x^2}{16}}-{\displaystyle \frac{y^2}{b^2}}=1$

Eccentricity:

$$e_2=\sqrt{1+\frac{b^2}{16}}\mathrm{.<span\; class="mord\; sqrt"><br></span>}$$

Given $e_1 e_2 = 1$:

$$\sqrt{1-\frac{b^2}{25}}\sqrt{1+\frac{b^2}{16}}=1.$$

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Step 2 — Solve for $b$

Square both sides:

$$(1-\frac{b^2}{25})(1+\frac{b^2}{16})=1.$$

Expand:

$$1+\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=1.$$

Simplify:

$$\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=0.$$

Multiply by 400:

$$25 b^2 - 16 b^2 - b^4 = 0$$
$$9b^2-b^4=0\Rightarrow b^2(9-b^2)=0.$$

So $b^2 = 9$ (since $b\neq0$).
With $b<5$, we get $b = 3$.

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Step 3 — Coordinates of all foci

Ellipse foci ($y$-axis major):

$$c_1=\sqrt{25-b^2}=\sqrt{25-9}=4.$$

Foci: $(0, \pm 4)$.

Hyperbola foci (along $x$-axis):

$$c_2=\sqrt{16+b^2}=\sqrt{16+9}=5.$$

Foci: $(\pm 5, 0)$.

So four foci: $(0,4), (0,-4), (5,0), (-5,0)$

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Step 4 — Ellipse through the four foci

Let required ellipse:

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

(since axes along coordinates).

Plug $(5,0)$: $25/A^2 = 1 \Rightarrow A^2 = 25$.
Plug $(0,4):$$16/B^2 = 1 \Rightarrow B^2 = 16$.

So ellipse is:

$$\frac{x^2}{25}+\frac{y^2}{16}=1.$$

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Step 5 — Eccentricity of required ellipse

Major axis along $x$-axis ($A=5$).
Eccentricity:

$$e=\sqrt{1-\frac{B^2}{A^2}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\mathrm{.}$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

The ellipse passing through all four foci is:

$$\frac{x^2}{25}+\frac{y^2}{16}=1$$

and its eccentricity is:

$$\boxed{{\displaystyle e=\frac{3}{\mathrm{5<span\; class="vlist-s"></span>}}}}$$

▶️ Watch the detailed explanation here: