If $y - 2 x - k = 0$ touches the conic $3 x^{2} - 5 y^{2} = 15$ , find the value of $k$ .

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Solution

$y = 2 x + k$
Putting the value of $y$ in the conic

$3 x^{2} - 5 {(2 x + k)}^{2} - 15 = 0$

$3 x^{2} - 5 (4 x^{2} + k^{2} + 4 x k) - 15 = 0$

$3 x^{2} - 20 x^{2} - 5 k^{2} - 20 x k - 15 = 0$

$17 x^{2} + 20 k x + 5 k^{2} + 15 = 0$

Upon solving the quadratic equation , for having a real value $D \geq 0$

$D = b^{2} - 4 a c$

$D = 400 k^{2} - 68 (5 k^{2} + 15)$

$D = 60 (k^{2} - 17)$

$D \geq 0$
$60 (k^{2} - 17) \geq 0$

$k^{2} - 17 \geq 0$

$k^{2} \geq 17$

$- \sqrt{1} 7 \leq k \leq \sqrt{1} 7$

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