Respuesta :
Answer:
The equations of the two tangents are [tex]y=12x[/tex] and [tex]y=-12x[/tex].
Step-by-step explanation:
The given curve is
[tex]y=2x^2+18[/tex] .... (1)
Let the point of tangency is at (a,b).
[tex]b=2a^2+18[/tex] .... (2)
Differentiate (1) with respect to x.
[tex]\frac{dy}{dx}=2(2x)+(0)[/tex]
[tex]\frac{dy}{dx}=4x[/tex]
[tex]\frac{dy}{dx}_{(a,b)}=4a[/tex]
The slope of tangent is 4a.
It is given that tangent passes through the point (a,b) with slope 4a. So, equation of tangent is
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope.
[tex]y-b=4a(x-a)[/tex] ... (3)
The line passes through the point (0,0).
[tex]0-b=4a(0-a)[/tex]
[tex]-b=4a(-a)[/tex]
[tex]b=4a^2[/tex] .... (4)
From (1) and (4) we get
[tex]4a^2=2a^2+18[/tex]
[tex]4a^2-2a^2=18[/tex]
[tex]2a^2=18[/tex]
[tex]a^2=9[/tex]
Taking square root on both sides.
[tex]a=\pm 3[/tex]
Substitute [tex]a^2=9[/tex] in equation (4).
[tex]b=4(9)=36[/tex]
The points of tangency are (3,36) and (-3,36).
Substitute the value of a and b in equation (3) to find the equations of tangents.
For (3,36),
[tex]y-36=4(3)(x-3)[/tex]
[tex]y-36=12x-36[/tex]
[tex]y=12x[/tex]
For (-3,36),
[tex]y-36=4(-3)(x-(-3))[/tex]
[tex]y-36=-12x-36[/tex]
[tex]y=-12x[/tex]
Therefore, the equations of the two tangents are [tex]y=12x[/tex] and [tex]y=-12x[/tex].
