Question

a line has equation 2x+y=20 and the curve has equation y=a+18/x-3, where a is a constant

find the set of values of A for which the line does not intersect the curve.

Answer 1

Answer:

For $a\in (2,26)$, the line does not intersect the curve

Step-by-step explanation:

Given the equation of the line

$2x+y=20$

The curve given is

$y=a+18/(x-3)$

If the line intersects the curve then

$2x+a+\frac{18}{x-3}=20$

$\implies 2x(x-3)+18=(20-a)(x-3)$

$\implies 2x^2-6x+18=(20-a)x-60+3a$

$\implies 2x^2-(26-a)x+(78-3a)=0$

For the line and curve not to intersect, the discriminant of the above quadratic equation should be less than zero

Therefore,

$(26-a)^2-4\times 2\times (78-3a)<0$

$\implies 676-52a+a^2-624+24a<0$

$\implies 52-28a+a^2<0$

$\implies a^2-28a+52<0$

$\implies a^2-26a-2a+52<0$

$\implies a(a-26)-2(a-26)<0$

$\implies (a-2)(a-26)<0$

$\implies a\in (2,26)$

Therefore for $a\in (2,26)$, the line does not intersect the curve.

Hope this helps.

Answer 2

$\huge\bf{Answer:-}$

Refer the attachment.