A) \[circle\,{{x}^{2}}+{{y}^{2}}=3\]
B) \[line\,y=\frac{\sqrt{3}}{4}\]
C) \[line\,y=\frac{\sqrt{3}}{4}\]
D) \[parabola\,{{y}^{2}}=\frac{3x}{-16}\]
Correct Answer: C
Solution :
Let f(x) has degree n so\[{{n}^{2}}=1+n+1\Rightarrow n=2\] \[\Rightarrow \]f(x) is quadratic with\[f(0)=0\] so let\[f(x)=a{{x}^{2}}+bx\] so\[a{{(a{{x}^{2}}+bx)}^{2}}+b(a{{x}^{2}}+bx)=x\left[ \frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2} \right]\forall x\in R\]\[\Rightarrow \]\[{{a}^{3}}-\frac{a}{3}=0;2{{a}^{2}}b-\frac{b}{2}=0;a{{b}^{2}}+ab=0\]and\[{{b}^{2}}=0\]\[\Rightarrow \]b = 0 and \[a=\pm \frac{1}{\sqrt{3}}\] \[\therefore \]given leading coefficient is positive \[\Rightarrow \]\[f(x)=\frac{{{x}^{2}}}{\sqrt{3}}\] So required curve is parabola Hence perpendicular tangents intersect on directrix. Chord of contact of any point on directrix always passes through focus.
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