JEE Main & Advanced
Sample Paper
JEE Main - Mock Test - 26
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.