Question

Find the equations of the two tangents to the curve y= 2x2 + 18 which pass through the origin.

Answer 1

Answer:

The equations of the two tangents are $y=12x$ and $y=-12x$.

Step-by-step explanation:

The given curve is

$y=2x^2+18$           .... (1)

Let the point of tangency is at (a,b).

$b=2a^2+18$             .... (2)

Differentiate (1) with respect to x.

$\frac{dy}{dx}=2(2x)+(0)$

$\frac{dy}{dx}=4x$

$\frac{dy}{dx}_{(a,b)}=4a$

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

$y-y_1=m(x-x_1)$

where, m is slope.

$y-b=4a(x-a)$         ... (3)

The line passes through the point (0,0).

$0-b=4a(0-a)$

$-b=4a(-a)$

$b=4a^2$             .... (4)

From (1) and (4) we get

$4a^2=2a^2+18$

$4a^2-2a^2=18$

$2a^2=18$

$a^2=9$

Taking square root on both sides.

$a=\pm 3$

Substitute $a^2=9$ in equation (4).

$b=4(9)=36$

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),

$y-36=4(3)(x-3)$

$y-36=12x-36$

$y=12x$

For (-3,36),

$y-36=4(-3)(x-(-3))$

$y-36=-12x-36$

$y=-12x$

Therefore, the equations of the two tangents are $y=12x$ and $y=-12x$.