question_answer

For a variable ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\] if distance between focus and extremity of minor axis is fixed, then locus of one end of its latus rectum will be

A) Straight line        

B)       Circle                  

C)    Parabola       

D)        Ellipse

Correct Answer: C

Solution :

[c]             \[\sqrt{{{a}^{2}}{{e}^{2}}+{{b}^{2}}}=c\]             (constant) \[\Rightarrow \,\,\,\,\sqrt{{{a}^{2}}-{{b}^{2}}+{{b}^{2}}}=c\] \[\Rightarrow \,\,\,\,\,\,\,c=a\]                               ??(1) Now,  \[h=ae=\sqrt{{{a}^{2}}-{{b}^{2}}}\]b          ?....(2) and \[k=\frac{{{b}^{2}}}{a}\]                            ?...(3) Eliminating b from (2) and (3), we get \[h=\sqrt{{{a}^{2}}-ka}\] Putting \[a=c,\]we get \[{{h}^{2}}={{c}^{2}}-kc\] Thus, the required locus is \[{{x}^{2}}={{c}^{2}}-cy,\]which is parabola.