Second part again: The point (h,k) lies on the curve y=2x²+18.Find the gradient at...

[Kitimbo]

Second part again: The point (h,k) lies on the curve y=2x²+18.Find the gradient at...

The point (h,k) lies on the curve y=2x²+18.Find the gradient at this point and the equation of the tangent there.Hence find the equations of the two tangents to the curve which pass through the origin.

So I did the first part to get grad y=4h and the equation of the tangent y-k=4h(x-h).
Help me with second part

 

[Harry_the_cat]

The point (h,k) lies on the curve y=2x²+18.Find the gradient at this point and the equation of the tangent there.Hence find the equations of the two tangents to the curve which pass through the origin.

So I did the first part to get grad y=4h and the equation of the tangent y-k=4h(x-h).
Help me with second part please

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[HallsofIvy]

The point (h,k) lies on the curve y=2x²+18.Find the gradient at this point and the equation of the tangent there.Hence find the equations of the two tangents to the curve which pass through the origin.

So I did the first part to get grad y=4h and the equation of the tangent y-k=4h(x-h).
Help me with second part

If a line given by y- k= 4h(x- h) passes through the origin, (0, 0), then x= 0, y= 0 satisfies that equation!
So you must have 0- k= 4h(0- h) or -k= -4h². Of course, since you are told that (h, k) lies on the curve, you must also have k=2h²+18. You have two equations to solve for h and k.

 

[JeffM]

The point (h,k) lies on the curve y=2x²+18.Find the gradient at this point and the equation of the tangent there.Hence find the equations of the two tangents to the curve which pass through the origin.

So I did the first part to get grad y=4h and the equation of the tangent y-k=4h(x-h).
Help me with second part

I shall build on Halls of Ivy's preceding post.

You did not go as far as you could with part 1 of the problem. Let's start with some notation.

\(\displaystyle f(x) = 2x^2 + 18 \implies f'(x) = 4x.\)

\(\displaystyle (h,\ k)\ is\ a\ point\ on\ f(x) \implies k = f(h) = 2h^2 + 18 \implies f'(h) = 4h.\)

\(\displaystyle equation\ of\ tangent\ line\ when\ x\ = h\ is\ g_h(x) \implies\)

\(\displaystyle g_h(x) = u + vx\ and\ derivative\ of\ g_h(x) = v.\)

\(\displaystyle But\ derivative\ of\ g_h(h) = f'(h) = 4h \implies v = 4h \implies g_h(x) = u + 4hx.\)

\(\displaystyle And g_h(h) = f(h) = 2h^2 + 18 = u + 4h(h) \implies u = 18 - 2h^2.\)

\(\displaystyle THUS\ g_h(x) = 18 - 2h^2 + 4hx.\) This is the completed answer to part 1.

Now, as Halls said. the equation of any line through the origin is

\(\displaystyle e(x) = wx = 0 + wx.\)

\(\displaystyle \therefore 0 + wx = g_h(x) = 18 - 2h^2 + 4hx \implies w = 4h\ and\ 18 - 2h^2 = 0 \implies \)

\(\displaystyle h = \pm 3 \implies g_{-3}(x) = (-\ 3)^2 + 18 + 4(-\ 3)x = 36 - 12x\ and\)

\(\displaystyle g_3(x) = 3^2 + 18 + 4(3)x = 36 + 12x.\)