Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16 x$. If one of these tangents touches the two curves at $Q$ and $R$, then $(Q R)^2$ is equal to :
Solution
<p>Let a tangent on ${y^2} = 16x$ be $y = mx + {4 \over m}$</p>
<p>For common to ${x^2} + {y^2} = 8$</p>
<p>${4 \over m} = 2\sqrt 2 (1 + {m^2})$</p>
<p>$\Rightarrow {2 \over {{m^2}}} = 1 + {m^2} \Rightarrow m = \, \pm 1$</p>
<p>Taking one of the tangent $y = x + 4$</p>
<p>Point of tangency with ${y^2} = 4x$</p>
<p>${x^2} + 8 + 16 = 4x \Rightarrow x = 4$ & $y = 8$</p>
<p>$\therefore$ $Q(4,8)$</p>
<p>and for ${x^2} + {y^2} = 8$</p>
<p>$2{x^2} + 8x + 8 = 0$</p>
<p>${x^2} + 4x + 4 = 8 \Rightarrow x = - 2,y = 2 \Rightarrow R = ( - 2,2)$</p>
<p>${(QR)^2} = {6^2} + {6^2} = 72$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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