An ellipse intersects the hyperbola 2x^2 - 2y^2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes then

equation of ellipse is x^2 + 2y^2 = 2, the foci of ellipse are ( \pm 1,0 )

equation of ellipse is x^2 + 2y^2 = 2

the foci of ellipse are ( \pm 1,0 )

equation of ellipse is x^2 + 2y^2 = 4

the foci of ellipse are ( \pm \sqrt 2 ,0 )

Ellipse PYQ — JEE Advanced 2009 · Shift 2 · Q36 with Solution

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Correct answer: A, B

Given $2x^2-2y^2=1 \implies x^2-y^2=\frac12.$

So, $\frac{x^2}{1/2}-\frac{y^2}{1/2}=1.$

Hence for the hyperbola, $a_h^2=b_h^2=\frac12.$

Its eccentricity is $e_h=\sqrt{1+\frac{b_h^2}{a_h^2}}=\sqrt{1+1}=\sqrt2.$

Therefore the ellipse has eccentricity reciprocal to this: $e=\frac{1}{\sqrt2}.$

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b.$

Its eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}=\frac{1}{\sqrt2}.$

So,

\implies \frac{b^2}{a^2}=\frac12 \implies b^2=\frac{a^2}{2}.$$ --- 3. **Use orthogonality of intersection** At a common point of the two curves, their tangents are perpendicular, so product of slopes is $-1$. ### Slope of hyperbola Differentiate $$2x^2-2y^2=1$$ with respect to $x$: $$4x-4y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=\frac{x}{y}.$$ ### Slope of ellipse Differentiate $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ with respect to $x$: $$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{b^2x}{a^2y}.$$ Orthogonality gives $$\frac{x}{y}\left(-\frac{b^2x}{a^2y}\right)=-1.$$ So, $$\frac{b^2x^2}{a^2y^2}=1 \implies b^2x^2=a^2y^2.

Using $b^2=\frac{a^2}{2}$ ,

\implies y^2=\frac{x^2}{2}.$$ --- 4. **Use the fact that the common point lies on the hyperbola** From hyperbola: $$x^2-y^2=\frac12.$$ Substitute $y^2=\frac{x^2}{2}$: $$x^2-\frac{x^2}{2}=\frac12 \implies \frac{x^2}{2}=\frac12 \implies x^2=1,

thus $y^2=\frac12.$

So the intersection points are $(\pm 1, \pm \tfrac{1}{\sqrt2})$ with matching signs as applicable.

Since $(1,\frac1{\sqrt2})$ lies on the ellipse, $\frac{1}{a^2}+\frac{1/2}{b^2}=1.$ Using $b^2=\frac{a^2}{2}$ ,

\implies \frac{1}{a^2}+\frac{1}{a^2}=1 \implies \frac{2}{a^2}=1 \implies a^2=2.$$ Then $$b^2=\frac{a^2}{2}=1.$$ Hence the ellipse is $$\frac{x^2}{2}+y^2=1,$$ which is equivalent to $$x^2+2y^2=2.$$ So **Option A is correct**. --- 6. **Find the foci of the ellipse** For the ellipse, $$c^2=a^2-b^2=2-1=1 \implies c=1.$$ Therefore foci are $$(\pm c,0)=(\pm 1,0).$$ So **Option B is correct**. --- 7. **Check remaining options** - **Option C:** $x^2+2y^2=4$ means $$\frac{x^2}{4}+\frac{y^2}{2}=1,$$ so $a^2=4, b^2=2$, giving $$e=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt2},$$ but it does **not** satisfy the orthogonal intersection condition with the given hyperbola. Hence false. - **Option D:** foci $(\pm\sqrt2,0)$ would require $$c=\sqrt2,$$ but we found $c=1$. Hence false. --- 8. **Final answer** The correct options are: $$\boxed{A,\ B}$$ This matches the stored correct answer.