13. At any point $P$ on the parabola $y ^ { 2 } - 2 y - 4 x + 5 = 0$, a tangent is drawn which meets the directrix at $Q$. Find the locus of point R which divides QP externally in the ratio $\frac { 1 } { 2 } : 1$. Sol. Any point on the parabola is $\mathrm { P } \left( 1 + \mathrm { t } ^ { 2 } , 1 + 2 \mathrm { t } \right)$. The equation of the tangent at P is $\mathrm { t } ( \mathrm { y } - 1 ) = \mathrm { x } - 1 + \mathrm { t } ^ { 2 }$ which meets the directrix $\mathrm { x } = 0$ at $\mathrm { Q } \left( 0,1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } \right)$. Let R be $( \mathrm { h } , \mathrm { k } )$. Since it divides QP externally in the ratio $\frac { 1 } { 2 } : 1 , \mathrm { Q }$ is the mid point of RP $\Rightarrow 0 = \frac { \mathrm { h } + 1 + \mathrm { t } ^ { 2 } } { 2 }$ or $\mathrm { t } ^ { 2 } = - ( \mathrm { h } + 1 )$ and $1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } = \frac { \mathrm { k } + 1 + 2 \mathrm { t } } { 2 }$ or $\mathrm { t } = \frac { 2 } { 1 - \mathrm { k } }$ So that $\frac { 4 } { ( 1 - k ) ^ { 2 } } + ( h + 1 ) = 0$ Or $( k - 1 ) ^ { 2 } ( h + 1 ) + 4 = 0$. Hence locus is $( y - 1 ) ^ { 2 } ( x + 1 ) + 4 = 0$.
The area of the triangle formed by the line $x + y = 3$ and the angle bisectors of the pair of straight lines $x ^ { 2 } - y ^ { 2 } + 2 y = 1$ is
13. At any point $P$ on the parabola $y ^ { 2 } - 2 y - 4 x + 5 = 0$, a tangent is drawn which meets the directrix at $Q$. Find the locus of point R which divides QP externally in the ratio $\frac { 1 } { 2 } : 1$.
Sol. Any point on the parabola is $\mathrm { P } \left( 1 + \mathrm { t } ^ { 2 } , 1 + 2 \mathrm { t } \right)$. The equation of the tangent at P is $\mathrm { t } ( \mathrm { y } - 1 ) = \mathrm { x } - 1 + \mathrm { t } ^ { 2 }$ which meets the directrix $\mathrm { x } = 0$ at $\mathrm { Q } \left( 0,1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } \right)$. Let R be $( \mathrm { h } , \mathrm { k } )$.\\
Since it divides QP externally in the ratio $\frac { 1 } { 2 } : 1 , \mathrm { Q }$ is the mid point of RP\\
$\Rightarrow 0 = \frac { \mathrm { h } + 1 + \mathrm { t } ^ { 2 } } { 2 }$ or $\mathrm { t } ^ { 2 } = - ( \mathrm { h } + 1 )$\\
and $1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } = \frac { \mathrm { k } + 1 + 2 \mathrm { t } } { 2 }$ or $\mathrm { t } = \frac { 2 } { 1 - \mathrm { k } }$\\
So that $\frac { 4 } { ( 1 - k ) ^ { 2 } } + ( h + 1 ) = 0$ Or $( k - 1 ) ^ { 2 } ( h + 1 ) + 4 = 0$.\\
Hence locus is $( y - 1 ) ^ { 2 } ( x + 1 ) + 4 = 0$.\\