Help on Parametrics! (1 Viewer)

[Heresy]

I am currently having trouble with part b - any help would be appreciated.

7a) Find where the line y= 3x + 4 intersects the parabola 2y=5x^2
Answers: (2,10) and (-4/5, 8/5)

7b) Find the equations of the tangents to the parabola at the points of intersection.

 

[mariakery1]

Dont u just differentiate the equation of the parabola, find the gradients at x=2 and x=-4/5 and sub into y-y1=m(x-x1)? i start parametrics next term so maybe im wrong

 

[jathu123]

Find the derivative of the parabola at each poiints and then use the point-gradient formula to find the eqn of tangents

y = 5/2 x^2

dy/dx = 5x

at x = 2, dy/dx = 10
therefore, the eqn of tangent at (2,10) is y - 10 = 10(x-2) [rearrange into the general form]

at x = -4/5, dy/dx = -4
therefore, the eqn of tangent at (-4/5, 8/5) is y - 8/5 = -4(x + 4/5) [again, rearrrange it]

 

[Heresy]

Find the derivative of the parabola at each poiints and then use the point-gradient formula to find the eqn of tangents

y = 5/2 x^2

dy/dx = 5x

at x = 2, dy/dx = 10
therefore, the eqn of tangent at (2,10) is y - 10 = 10(x-2) [rearrange into the general form]

at x = -4/5, dy/dx = -4
therefore, the eqn of tangent at (-4/5, 8/5) is y - 8/5 = -4(x + 4/5) [again, rearrrange it]

Thanks!