C.

$\sqrt{1 - 2\mathrm{a} +2{\mathrm{a}}^2}$

$$\begin{gathered} \mathrm{Given} \mathrm{curve} \mathrm{is} \\ 2{\mathrm{x}}^2 + {\mathrm{y}}^2 = 2\mathrm{x} \\ 2{\mathrm{x}}^2 - 2\mathrm{x} + {\mathrm{y}}^2 = 0 \\ \Rightarrow 2{\left(\mathrm{x} - \frac{1}{2}\right)}^2 +\hspace{0.17em}{\mathrm{y}}^2 = \frac{1}{2} \\ \Rightarrow \frac{{\left(\mathrm{x} - \frac{1}{2}\right)}^2 }{{\displaystyle \frac{1}{4}}} + \frac{{\displaystyle \frac{{\mathrm{y}}^2}{1}}}{2} = 1 \\ \mathrm{which} \mathrm{represents} \mathrm{an} \mathrm{ellipse} \\ \mathrm{Here}, \mathrm{a} = \frac{1}{2}, \mathrm{b} = \frac{1}{\sqrt{2}}, \mathrm{h} = \frac{1}{2}, \mathrm{k} = 0 \\ \mathrm{Consider} \mathrm{a} \mathrm{point} \mathrm{P}(\mathrm{h} + \mathrm{acos}\left(\mathrm{\theta }\right), \mathrm{k} + \mathrm{bsin}\left(\mathrm{\theta }\right)) \\ = \mathrm{P}\left(\frac{1}{2} + \frac{1}{2}\cos \left(\mathrm{\theta }\right), \frac{1}{\sqrt{2}}\sin \left(\mathrm{\theta }\right)\right) \mathrm{on} \mathrm{the} \mathrm{ellipse} \mathrm{from} \mathrm{which} \mathrm{the} \mathrm{distance} \\ \mathrm{of} \mathrm{point} (\mathrm{a}, 0) \mathrm{is} \mathrm{maximum} \\ \mathrm{let} \mathrm{Q}(\mathrm{a}, 0) \\ \mathrm{Now}, \\ \mathrm{PQ} = \sqrt{{\left(\frac{1}{2} + \frac{1}{2}\cos \left(\mathrm{\theta }\right) - \mathrm{a}\right)}^2 + {\left(\frac{1}{\sqrt{2}}\sin \left(\mathrm{\theta }\right) - 0\right)}^2 } \\ = \sqrt{\frac{1}{4} + \frac{1}{4}{\cos }^2\left(\mathrm{\theta }\right) + {\mathrm{a}}^2 + \frac{1}{2}\cos \left(\mathrm{\theta }\right) - \mathrm{acos}\left(\mathrm{\theta }\right) - \mathrm{a} + \frac{1}{2}{\sin }^2\left(\mathrm{\theta }\right)} \\ \mathrm{PQ} = \sqrt{\frac{1}{2} + {\mathrm{a}}^2 - \mathrm{a} + \left(\frac{1}{2} - \mathrm{a}\right)\cos \left(\mathrm{\theta }\right) + \frac{1}{4}{\sin }^2\left(\mathrm{\theta }\right)} \\ \mathrm{Let} \mathrm{y} = {\mathrm{PQ}}^2 = \frac{1}{2} + {\mathrm{a}}^2 - \mathrm{a} + \left(\frac{1}{2} - \mathrm{a}\right)\cos \left(\mathrm{\theta }\right) + \frac{1}{4}{\sin }^2\left(\mathrm{\theta }\right) \end{gathered}$$

$$\begin{gathered} \mathrm{For} \mathrm{maxima} \mathrm{and} \mathrm{minima}, \mathrm{put} \frac{\mathrm{dy}}{\mathrm{d\theta }} = 0 \\ \Rightarrow - \left(\frac{1}{2} - \mathrm{a}\right)\sin \left(\mathrm{\theta }\right) + \frac{1}{4} . 2\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right) = 0 \\ \Rightarrow \sin \left(\mathrm{\theta }\right)\left(- \frac{1}{2} + \mathrm{a} + \frac{1}{2}\cos \left(\mathrm{\theta }\right)\right) = 0 \\ \Rightarrow \sin \left(\mathrm{\theta }\right) = 0 \mathrm{or} - \frac{1}{2} + \mathrm{a} + \frac{1}{2}\cos \left(\mathrm{\theta }\right) = 0 \\ \Rightarrow \mathrm{\theta } = 0 \mathrm{or} \cos \left(\mathrm{\theta }\right) = 1 - 2\mathrm{a} \\ \Rightarrow {\sin }^2\left(\mathrm{\theta }\right) = 1 - {\cos }^2\left(\mathrm{\theta }\right) \\ = 1 - {\left(1 - 2\mathrm{a}\right)}^2 \\ = - 4{\mathrm{a}}^2 + 4\mathrm{a} \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{d\theta }}^2} < 0 \mathrm{for} \cos \left(\mathrm{\theta }\right) = 1 - 2\mathrm{a} \\ \mathrm{Thus}, \mathrm{distance} \mathrm{PQ} \mathrm{is} \mathrm{maximum}, \mathrm{when} \\ \cos \left(\mathrm{\theta }\right) = 1 - 2\mathrm{a} \mathrm{and} {\sin }^2\left(\mathrm{\theta }\right) = - 4{\mathrm{a}}^2 + 4\mathrm{a} \\ \mathrm{Now}, \mathrm{required} \mathrm{longest} \mathrm{distance} \mathrm{is} \\ =\sqrt{\frac{1}{2} + {\mathrm{a}}^2 - \mathrm{a} + \left(\frac{1}{2} - \mathrm{a}\right)\left(1 - 2\mathrm{a}\right) + \frac{1}{4}\left(- 4{\mathrm{a}}^2 + 4\mathrm{a}\right)} \\ = \sqrt{\frac{1}{2} + {\mathrm{a}}^2 - \mathrm{a} + \frac{1}{2} - \mathrm{a} - \mathrm{a} + 2{\mathrm{a}}^2 - {\mathrm{a}}^2 + \mathrm{a}} \\ = \sqrt{2{\mathrm{a}}^2 - 2\mathrm{a} + 1} \\ = \sqrt{1 - 2\mathrm{a} + 2{\mathrm{a}}^2} \end{gathered}$$

A.

$\left(\frac{5\mathrm{a} +\hspace{0.17em}\mathrm{d}}{6}, \frac{\mathrm{b} + 5\mathrm{c}}{6}\right)$

$$\begin{gathered} \mathrm{Centroid} \mathrm{of} ∆\mathrm{ABC} = \left(\frac{\mathrm{a} +\mathrm{a} +\hspace{0.17em}\mathrm{d}}{3}, \frac{\mathrm{b} + \mathrm{c} + \mathrm{c}}{3}\right) \\ \equiv \left(\frac{2\mathrm{a} + \mathrm{d}}{3}, \frac{2\mathrm{c} + \mathrm{b}}{3}\right) \\ \mathrm{Since}, ∆\mathrm{ABC} \mathrm{is} \left(\mathrm{a}, \mathrm{c}\right) \\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{centroid} \mathrm{and} \mathrm{orthocentre} \mathrm{is} \\ = \left(\frac{\frac{2\mathrm{a} + \mathrm{d}}{3} + \mathrm{a}}{2}, \frac{\frac{2\mathrm{c} + \mathrm{b}}{3} + \mathrm{c}}{2}\right) \\ = \left(\frac{5\mathrm{a} + \mathrm{d}}{6}, \frac{5\mathrm{c} + \mathrm{b}}{6}\right) \end{gathered}$$

A.

(0, 2)

$\left(\mathrm{a}\right) \mathrm{The} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{given} \mathrm{lines} \mathrm{is} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{adjacent} \mathrm{figure}$

Clearly, the tnangle ABC is an isosceles triangle

$\therefore$ The incentre he on the median to the base

$\because$ D is mid-point of BC

$\therefore$ OD is median to the base BC

Thus, incentre lie on Y-axis

A.

$\left(\frac{4}{3}, \frac{4}{3}\right)$

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{x} + \mathrm{y} = 1 \\ \mathrm{and} 2{\mathrm{y}}^2 - \mathrm{xy} - 6{\mathrm{x}}^2 = 0 \\ \Rightarrow 2{\mathrm{y}}^2 - 4\mathrm{xy} + 3\mathrm{xy} - 6{\mathrm{x}}^2 = 0 \\ \Rightarrow \mathrm{Equation} \mathrm{of} \mathrm{sides} \mathrm{of} ∆\mathrm{ABC} \mathrm{are} \\ \mathrm{x} + \mathrm{y} = 1, 2\mathrm{y} + 3\mathrm{x} = 0 \mathrm{and} \mathrm{y} - 2\mathrm{x} = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{Solving} \mathrm{these} \mathrm{equations} \mathrm{simultaneously}, \mathrm{we} \mathrm{get} \mathrm{A}\left(0, 0\right) \\ \mathrm{B}\left(\frac{1}{3}, \frac{2}{3}\right) \mathrm{and} \mathrm{C}\left( - 2, 3\right) \\ \mathrm{Equation} \mathrm{of} \mathrm{altitude} \mathrm{AD} \mathrm{is} \\ \mathrm{x} - \mathrm{y} = 0 ... \left(\mathrm{i}\right) \\ \mathrm{Equation} \mathrm{of} \mathrm{altitude} \mathrm{CF} \mathrm{is} \\ \mathrm{x} + 2\mathrm{y} = \mathrm{\lambda } \\ \mathrm{Since}, \mathrm{this} \mathrm{passes} \mathrm{through} \left( - 2, 3\right) \\ \therefore - 2 + 6 = \mathrm{\lambda } \Rightarrow \mathrm{\lambda } = 4 \\ \mathrm{On} \mathrm{solving} \mathrm{eqs}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ \mathrm{x} = \frac{4}{3}, \mathrm{y} = \frac{4}{3} \end{gathered}$$

B.

1

$$\begin{gathered} \mathrm{We} \mathrm{have}, \\ \mathrm{xy} - \mathrm{x} - \mathrm{y} + 1 = 0 ...\left(\mathrm{i}\right) \\ \mathrm{ax} + 2\mathrm{y} - 3\mathrm{a} = 0 ...\left(\mathrm{ii}\right) \\ \mathrm{On} \mathrm{compairing} \mathrm{Eqs}. \left(\mathrm{i}\right) \mathrm{with}, \\ {\mathrm{ax}}^2 + 2\mathrm{hxy} + {\mathrm{by}}^2 + 2\mathrm{gx} + 2\mathrm{fy} + \mathrm{c} = 0, \mathrm{we} \mathrm{get} \\ \mathrm{a} = 0, \mathrm{b} = 0, \mathrm{h} = \frac{1}{2}, \mathrm{g} = - \frac{1}{2}, \mathrm{f} = - \frac{1}{2}, \mathrm{c}=1 \\ \mathrm{The} \mathrm{point} \mathrm{of} \mathrm{concurrent} \mathrm{is} \left(\frac{\mathrm{hf} - \mathrm{bg}}{\mathrm{ab} - {\mathrm{h}}^2}, \frac{\mathrm{gh} - \mathrm{af}}{\mathrm{ab} - {\mathrm{h}}^2}\right) \\ = \left(\frac{{\displaystyle \frac{1}{2}\left( - \frac{1}{2}\right) - 0}}{0 - {\left({\displaystyle \frac{1}{2}}\right)}^2}, \frac{\left( - \frac{1}{2}\right)\frac{1}{2} - 0}{0 - {\left(\frac{1}{2}\right)}^2}\right) = \left(1, 1\right) \\ \mathrm{The} \mathrm{point} \mathrm{lies} \mathrm{on} \mathrm{equation} \left(\mathrm{ii}\right), \mathrm{then} \\ \mathrm{a} + 2 - 3\mathrm{a} = 0 \Rightarrow 2\mathrm{a} = 2 \\ \mathrm{a} = 1 \end{gathered}$$