Consider the lines L_ 1 and L_ 2 defined by L_ 1 : x 2 + y-1=0 and...

From the first question

The equation of the locus is $\left|2{x}^{2} - ( y - 1{)}^{2}\right| = 27$

The line is $y = 2x + 1$ or $y - 1 = 2x$

By substituting the value of $y$ in the equation of the curve $C$, we get

\left|2{x}^{2} - ( y - 1{)}^{2}\right| = 27$</p><p>$⇒ \left|2{x}^{2} - ( 2x{)}^{2}\right| = 27$</p><p>$⇒ 2{x}^{2} = 27$</p><p>$⇒ x = \pm 3\sqrt{\dfrac{3}{2}}$</p><p>$⇒ {x}_{1}, {x}_{2} = \pm 3\sqrt{\dfrac{3}{2}}$</p><p>Let$M$be the mid-point of${R}^{'} \text{ and } {S}^{'}$</p><p>So, the$x$coordinate of$T$is$\dfrac{{x}_{1} + {x}_{2}}{2} = 0$</p><p>It lies on$y = 2x + 1$</p><p>So, the coordinates of$T$are \left(0, 1\right)$

Slope of the line perpendicular to $y = 2x + 1$ is $\dfrac{- 1}{2}$

So, the equation of perpendicular bisector is

$y - 1 = \dfrac{- 1}{2}\left(x - 0\right)$

Or, $x + 2y = 2$

Let coordinate of ${R}^{'} \text{ and } {S}^{'} are \left({p}_{1}, {q}_{1}\right) \text{ and } \left({p}_{2}, {q}_{2}\right)$, we get

$D = {\left({p}_{2} - {p}_{1}\right)}^{2} + {\left({q}_{2} - {q}_{1}\right)}^{2}$

Since both the points satisfy the equation of the line, we get

$D = {\left[2\left({q}_{2} - {q}_{1}\right)\right]}^{2} + {\left({q}_{2} - {q}_{1}\right)}^{2}$

$D = 5{\left({q}_{2} - {q}_{1}\right)}^{2}$

Solving, $x + 2y = 2$ with $2{x}^{2} - ( y - 1{)}^{2} = 27$, we get

$2{\left(2 - 2y\right)}^{2} - ( y - 1{)}^{2} = 27$

$⇒ 7 ( y - 1{)}^{2} = 27$

$⇒ y - 1 = \pm 3\sqrt{\dfrac{3}{7}}$

$⇒ y = 1 \pm 3\sqrt{\dfrac{3}{7}}$

$⇒ {q}_{1}, {q}_{2} = 1 \pm 3\sqrt{\dfrac{3}{7}}$

So, ${\left({q}_{2} - {q}_{1}\right)}^{2} = {\left(6\sqrt{\dfrac{3}{7}}\right)}^{2}$

Hence, $D = 5{\left({q}_{2} - {q}_{1}\right)}^{2} = 5 \times {\left(6\sqrt{\dfrac{3}{7}}\right)}^{2} = 77 . 14$