If the line $y = 4 + kx,\,k > 0$, is the tangent to the parabola $y = x - {x^2}$ at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :
Solution
<p>$\because$ Line $y = kx + 4$ touches the parabola $y = x - {x^2}$.</p>
<p>So, $kx + 4 = x - {x^2} \Rightarrow {x^2} + (k - 1)x + 4 = 0$ has only one root</p>
<p>${(k - 1)^2} = 16 \Rightarrow k = 5$ or $-$3 but $k > 0$</p>
<p>So, $k = 5$.</p>
<p>And hence ${x^2} + 4x + 4 = 0 \Rightarrow x = - 2$</p>
<p>So, P($-$2, $-$6) and V is $\left( {{1 \over 2},{2 \over 4}} \right)$</p>
<p>Slope of $PV = {{{1 \over 4} + 6} \over {{1 \over 2} + 2}} = {5 \over 2}$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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