A) centre only
B) centre, foci and directrices
C) centre, foci and vertices
D) centre and vertices only
Correct Answer: D
Solution :
Equation of ellipse is \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1,\] \[~a>b\]and equation of hyperbola is \[\frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{16}=1,\] \[a>b.\] Let e and e be the eccentricities of the ellipse and hyperbola. \[\therefore \] \[e=\sqrt{\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\frac{25-16}{25}}\] \[=\frac{3}{5}\] and \[e=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\frac{25+16}{25}}\] \[=\frac{\sqrt{41}}{5}\] (i) Centre of ellipse is \[(0,0)\] and centre of hyperbola is \[(0,0)\]. (ii) Foci of ellipse are \[(\pm \,ae,\,0)\] or \[(\pm \,5,\,0)\] foci of hyperbola are \[(\pm \,ae,\,0)\] or \[(\pm \,\sqrt{41}\,0,)\]. (iii) Directrices of ellipse are \[x=\pm \frac{a}{e}\] \[\Rightarrow \]\[x=\pm \frac{25}{3}\] directrices of hyperbola are \[x=\pm \frac{a}{e}\] \[\Rightarrow \] \[x=\pm \frac{25}{\sqrt{41}}\] (iv) Vertices of ellipse are \[(\pm \,a,0)\] or \[(\pm \,5,0)\] Vertices of hyperbola are \[(\pm \,a,0)\] or \[(\pm \,5,0)\] From the above discussions, their are common in centre and vertices.
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