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If the line $\mathrm{y=4+kx,k> 0}$ , is the tangent to the parabola $\mathrm{y=x-x^{2}}$ at the point $\mathrm{P}$ and $\mathrm{V}$ is the vertex of the parabola,then the slope of the line through $\mathrm{P}$ and $\mathrm{V}$ is :
Option: 1
$\frac{3}{2}$

Option: 2
$\frac{26}{9}$

Option: 3
$\frac{5}{2}$

Option: 4
$\frac{23}{6}$

Answers (1)

best_answer

$\mathrm{y=k x+4 }\\$

$\mathrm{y=x-x^{2}} \\$

$\mathrm{k x+4=x-x^{2}} \\$ .......(1)

$\mathrm{x^{2}+(k-1) x+4=0} \\$

$\mathrm{(k-1)^{2}-4 \cdot 4=0} \\$

$\mathrm{k-1=\pm 4 }\\$

if $\mathrm{ k=5}$

Now put the value of $\mathrm{ k=5}$

$\mathrm{5 x+4=x-x^{2}}$

$\mathrm{x^{2}+4 x+4=0} \\$

$\mathrm{(x+2)^{2}=0} \\$

$\mathrm{x=-2} \\$

$\mathrm {y=-6} \\$

if $\mathrm{k=-3}$

Now put the value of $\mathrm{k=-3}$ in eqn (1)

$\mathrm{-3 x+4=x-x^{2}}\\$

$\mathrm{x^{2}-4 x+4=0}\\$

$\mathrm{x=2 \quad y=-2}$

Then the point of $\mathrm{P}$ is $\mathrm{\left ( 2,-2 \right )}$ and $\mathrm{\left ( -2,-6 \right )}$ and vertex of parabola $\mathrm{'O'=y-\frac{1}{4}=-\frac{1}{4}+x-x^{2}}$

$\mathrm{y-\frac{1}{4}=-\left ( x-\frac{1}{2} \right )^{2}}$

Point $\mathrm{P}$ is $\mathrm{\left ( 2,2 \right )}$

Slope of $\mathrm{OP=\frac{-2-\frac{1}{4}}{2-\frac{1}{2}}=\frac{-3}{2}}$

Point $\mathrm{P}$ is $\mathrm{\left ( -2,-6 \right )}$ slope of $\mathrm{OP=\frac{-6-\frac{1}{4}}{-2-\frac{1}{2}}=\frac{5}{2}}$

Hence the correct answer is option 3

Posted by

Rakesh

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