There are two values of m where the line y=mx−1 is tangent to the parabola y=(x−2)^2+4. One of the possible values of m is positive, one is negative. How do I find the positive value of m?

Question

There are two values of m where the line y=mx−1 is tangent to the parabola y=(x−2)^2+4. One of the possible values of m is positive, one is negative. How do I find the positive value of m?

1 Answer

Impact of this question

7978 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-17T13:52:17

$m = 2$ $m=2$

Explanation:

Setting $y = m x - 1$ $y=mx-1$ in the equation of parabola: $y = {(x - 2)}^{2} + 4$ $y=(x-2)^2+4$ , we get

$m x - 1 = {(x - 2)}^{2} + 4$ $mx-1=(x-2)^2+4$

$m x - 1 = x^{2} - 4 x + 4 + 4$ $mx-1=x^2-4x+4+4$

$x^{2} - (m + 4) x + 9 = 0$ $x^2-(m+4)x+9=0$

Since, the given line: $y = m x - 1$ $y=mx-1$ is tangent to the parabola at a single point i.e. above equation must have equal real roots i.e. the determinant: $B^{2} - 4 A C$ $B^2-4AC$ of above quadratic equation must be zero as follows

${(- (m + 4))}^{2} - 4 (1) (9) = 0$ $(-(m+4))^2-4(1)(9)=0$

$m^{2} + 8 m + 16 - 36 = 0$ $m^2+8m+16-36=0$

$m^{2} + 8 m - 20 = 0$ $m^2+8m-20=0$

$m^{2} + 10 m - 2 m - 20 = 0$ $m^2+10m-2m-20=0$

$m (m + 10) - 2 (m + 10) = 0$ $m(m+10)-2(m+10)=0$

$(m + 10) (m - 2) = 0$ $(m+10)(m-2)=0$

$m+10=0\ \ or\ \ m-2=0$

$m=-10\ \ or\ \ m=2$

hence, the positive value of $m$ is $2$

Science

Math

Humanities

... and beyond

There are two values of m where the line y=mx−1 is tangent to the parabola y=(x−2)^2+4. One of the possible values of m is positive, one is negative. How do I find the positive value of m?

1 Answer

Explanation:

Related questions

Impact of this question