Mathematics - Ellipse Question with Solution | TestHub

Given, $2x^2-2y^2=1$ $$\Rightarrow \frac{x^2}{\left(\frac{1}{2}\right)}-\frac{y^2}{\left(\frac{1}{2}\right)}=1$$ Eccentricity of hyperbola $=\sqrt{2}$ So, eccentricity of ellipse $=1/\sqrt{2}$ Let equation of ellipse be $$\begin{array}{rl}& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)\Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}}\\ & \Rightarrow \frac{b^2}{a^2}=\frac{1}{2}\Rightarrow a^2=2b^2\\ & \therefore x^2+2y^2=2b^2\\ & \end{array}$$ Let ellipse and hyperbola intersect at $$A\left(\frac{1}{\sqrt{2}}\sec \theta ,\frac{1}{\sqrt{2}}\tan \theta \right)$$ On differentiating Eq. (i), $$4x-4y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{x}{y}$$ $$\therefore {\left(\frac{dy}{dx}\right)}_{\text{at }A}=\frac{\sec \theta }{\tan \theta }=\mathrm{cosec}\theta$$ On differentiating Eq. (ii), $$2x+4y\frac{dy}{dx}=0$$ $$\therefore {\left(\frac{dy}{dx}\right)}_{\text{at }A}=-\frac{x}{2y}=-\frac{1}{2}\mathrm{cosec}\theta$$ Since, ellipse and hyperbola are orthogonal. $$\begin{array}{rl}& \therefore -\frac{1}{2}{\mathrm{cosec}}^2\theta =-1\\ & \Rightarrow {\mathrm{cosec}}^2\theta =2\Rightarrow \theta =\pm \frac{\pi }{4}\\ & \therefore A\left(1,\frac{1}{\sqrt{2}}\right)\text{ or }\left(1,-\frac{1}{\sqrt{2}}\right)\end{array}$$ From Eq. (i), $1+2{\left(\frac{1}{\sqrt{2}}\right)}^2=2b^2$ $\Rightarrow b^2=1$ Equation of ellipse is $x^2+2y^2=2$ Coordinate of foci $(\pm ae,0)$ $$=\left(\pm \sqrt{2}\cdot \frac{1}{\sqrt{2}},0\right)=(\pm 1,0)$$ Hence, options (a) and (b) are correct. If major axis is along $Y$-axis, then $$\begin{array}{rl}& \frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}}\Rightarrow b^2=2a^2\\ & \therefore 2x^2+y^2=2a^2\Rightarrow y^{\prime }=-\frac{2x}{y}\\ & \Rightarrow y^{\prime }\left(\frac{1}{\sqrt{2}}\sec \theta \frac{1}{\sqrt{2}}\tan \theta \right)=\frac{-2}{\sin \theta }\\ & \end{array}$$ As ellipse and hyperbola are orthogonal. $$\begin{array}{ll}\therefore & -\frac{2}{\sin \theta }\cdot \mathrm{cosec}\theta =-1\\ \Rightarrow & {\mathrm{cosec}}^2\theta =1\Rightarrow \theta =\pm \frac{\pi }{4}\\ \therefore & 2x^2+y^2=2a^2\Rightarrow 2+\frac{1}{2}=2a^2\\ \Rightarrow & a^2=\frac{5}{4}\Rightarrow 2x^2+y^2=\frac{5}{2}\end{array}$$ Corresponding foci are $(0,\pm 1)$.