- A${30}$
- ✓${45}$
- C${60}$
- D${75}$
$\Rightarrow \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = 1$
$ \Rightarrow $ $\tan (\theta + \phi ) = 1$
$ \Rightarrow $ $\theta + \phi = \frac{\pi }{4} = 45^\circ$.
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$\begin{gathered}
f\left( x \right) = \left[ \begin{gathered}
{\cos ^{ - 1}}\left( \mu \right) + {x^2},0 < x < 1 \hfill \\
4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x \geqslant 1 \hfill \\
\end{gathered} \right.,f\left( x \right) \hfill \\
\hfill \\ \end{gathered}$ can have a local minimum at $x =$ $1$, if the value of $\mu$ lies in the interval
($1$)The orthocentre of the triangle $F_1 M N$ is
($A$) $\left(-\frac{9}{10}, 0\right)$ ($B$) $\left(\frac{2}{3}, 0\right)$ ($C$) $\left(\frac{9}{10}, 0\right)$ ($D$) $\left(\frac{2}{3}, \sqrt{6}\right)$
($2$) If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F_{\mathrm{I}} N F_2$ is
($A$) $3: 4$ ($B$) $4: 5$ ($C$) $5: 8$ ($D$) $2: 3$
Givan the answer qestion ($1$) and ($2$)