MCQ
The equation whose roots are $\frac{1}{{3 + \sqrt 2 }}$and $\frac{1}{{3 - \sqrt 2 }}$ is
- ✓$7{x^2} - 6x + 1 = 0$
- B$6{x^2} - 7x + 1 = 0$
- C${x^2} - 6x + 7 = 0$
- D${x^2} - 7x + 6 = 0$
${x^2} - \left( {\frac{1}{{3 + \sqrt 2 }} + \frac{1}{{3 - \sqrt 2 }}} \right)x + \frac{1}{{3 + \sqrt 2 }} \times \frac{1}{{3 - \sqrt 2 }} = 0$
$ \Rightarrow {x^2} - \left( {\frac{6}{7}} \right)x + \frac{1}{7} = 0 \Rightarrow 7{x^2} - 6x + 1 = 0$
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$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
| $List-I$ | $List-II$ |
| If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
| If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
| If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
| If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
| ($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is: