Question

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

Answer 1

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

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.

`(dy)/(dx)=2(2x)+(0)`

`(dy)/(dx)=4x`

`(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`.