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The value of k for which the line x+y+1=0 touches the parabola y^2=kx is
Dear studentPlease find the link to a similar problemhttps://www.askiitians.com/forums/Analytical-Geometry/if-the-line-x-y-1-0-touches-parabola-y-2-kx-then_224684.htmRegards
The line x+y=1 touches the parabola y^2=kx or y^2=4.k/4 x ( Compare with standard eqn y^2=4ax) so a= k/4 and the condition of tangency for the line y=mx+c to a parabola y^2=4ax is that : c=a/mwe ghave given c=1, m=-1, a=k/4 This implies 1=k/4(-1) => k=-4
x+y-1=0. or. y=1-x………(1)y^2. = kx……………………..(2)On putting y=1-x from eq.(1)(1-x)^2=k.x1–2x+x^2 =kxx^2 - (k+2)x+1 = 0The given line is tangent to the parabola which touches the parabola one and only one point. means both the roots are same.D or. b^ - 4 a.c. =0{-(k+2)}^2. - 4.1.1. =0(k+2)^2. = 4(k+2) = +/-(2)k= -2 +/-2k= -2+2 or. -2–2.k =0. , - 4. Answer.
x+y-1=0. or. y=1-x………(1)
y^2. = kx……………………..(2)
On putting y=1-x from eq.(1)
(1-x)^2=k.x
1–2x+x^2 =kx
x^2 - (k+2)x+1 = 0
The given line is tangent to the parabola which touches the parabola one and only one point. means both the roots are same.
D or. b^ - 4 a.c. =0
{-(k+2)}^2. - 4.1.1. =0
(k+2)^2. = 4
(k+2) = +/-(2)
k= -2 +/-2
k= -2+2 or. -2–2.
k =0. , - 4. Answer.
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