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At any point $\mathrm{P}$ on the parabola $\mathrm{y^2-2 y-4 x+5=0}$ a tangent is drawn which meets directrix at $\mathrm{Q}$ if the locus of point $\mathrm{R}$ Which divides $\mathrm{QP}$ externally in the rario $\mathrm{\frac{1}{2}:1}$ is $\mathrm{6 (x+a) ( y-a)^2+b=0}$ Find $\mathrm{a+b}$
Option: 1
$3$

Option: 2
$5$

Option: 3
$2$

Option: 4
$6$

Answers (1)

best_answer

Equation of the parabola is $\mathrm{(y-1)^2=4(x-1)}$

Let $\mathrm{P\left(t^2+1,2 t+1\right)}$

$\mathrm{ \frac{d y}{d x} \text { at } P=\frac{1}{t} }$

Equation of the tangent at $\mathrm{ P\: is \: x-y t+\left(t^2+t-1\right)=0 }$ Equation of the directrix is $\mathrm{ x=0 }$

$\mathrm{ \therefore Q\left(0, \frac{t^2+t-1}{t}\right) }$

$\mathrm{ R=(h, k)=\left(\frac{1+t^2-0}{-1}, \frac{2 t^2+t-2 t^2-2 t+2}{-t}\right) }$

$\mathrm{ \therefore h=-\left(1+t^2\right), k=\frac{t-2}{t}}$

Eliminating $\mathrm{ t,(h+1)(k-1)^2+4=0}$

Locus of $\mathrm{ R\: is \: (x+1)(y-1)^2+4=0.}$

Hence option 2 is correct.

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Shailly goel

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