Let a line L : 2x + y = k, k > 0 be a tangent to the hyperbola x2 $-$ y2 = 3. If L is also a tangent to the parabola y2 = $\alpha$x, then $\alpha$ is equal to :
Solution
Tangent to hyperbola of <br><br>Slope m = $-$2 (given)<br><br>y = $-$2x $\pm$ $\sqrt {3(3)}$<br><br>$\left( {y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} } \right)$<br><br>$\Rightarrow$ y + 2x = $\pm$ 3 $\Rightarrow$ 2x + y = 3 (k > 0)<br><br>For parabola y<sup>2</sup> = $\alpha$x<br><br>$y = mx + {\alpha \over {4m}}$<br><br>$\Rightarrow y = - 2x + {\alpha \over { - 8}}$<br><br>$\Rightarrow {\alpha \over { - 8}} = 3$<br><br>$\Rightarrow \alpha = - 24$
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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