## Differentiation tangent IB math HL

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### Differentiation tangent IB math HL

Differentiation, Derivatives, gradient, tangent - IB Mathematics HL

How can we find the equation of the tangent to the curve with equation

$2x^2+5xy-4y^2-22x=-60$ at the point $(5,0)$

Thanks
Jack

Posts: 0
Joined: Mon Jan 28, 2013 8:08 pm

### Re: Differentiation tangent IB math HL

Differentiation, Derivatives Equation of a tangent - IB maths HL

$2x^2+5xy-4y^2-22x=-60 \Rightarrow$

$\frac{d}{dx}{ 2x^2+5xy-4y^2-22x}=\frac{d}{dx}-60 \Rightarrow$

$4x+5(y+x \frac{dy}{dx})-8y \frac{dy}{dx}-22=0 \Rightarrow$

$4x+5y+5x \frac{dy}{dx}-8y \frac{dy}{dx}-22=0 \Rightarrow$

$\frac{dy}{dx}(5x-8y)=22-4x-5y \Rightarrow$

$\frac{dy}{dx}=\frac{22-4x-5y}{5x-8y}\Rightarrow$

The gradient of the curve at $(5,0)$ is

$\frac{dy}{dx}=\frac{22-4(5)-5(0)}{5(5)-8(0)}\Rightarrow$

$\frac{dy}{dx}=\frac{2}{25}$

The equation of the tangent on this point is given by the equation

$y-y_{0}=m_{t}(x-x_{0})$

where $m_{t}=\frac{2}{25}$ is the gradient of the tangent

and the touch point is $(x_{0}, y_{0})=(5,0)$.

So, the equation of the tangent at the given point is

$y=\frac{2}{25} (x-5)= \frac{2}{25} x-\frac{2}{5}$

Hope these help!!
elizabeth

Posts: 0
Joined: Mon Jan 28, 2013 8:09 pm

### Re: Differentiation tangent IB math HL

Thank you!!
Jack

Posts: 0
Joined: Mon Jan 28, 2013 8:08 pm