[Math] Equation of the tangent to $4x^2-3xy-y^2=25$ at $(2, -3)$

calculusimplicit-differentiationpolynomials

I have a function $f(x,y)=4x^2-3xy-y^2=25$ and I am trying to find the gradient at $(2,-3)$. First i took the derivative: $$\frac{d}{dx}(f'(x,y))=8x-\left(3y+3x\frac{dy}{dx}\right)-2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{-8x+3y}{-3x-2y}=\frac{-16-9}{{-6+6}}=\frac{-25}{0}$$

What is going wrong here? After I get the gradient right, do I put it in $y-y_1=m(x-x_1)$, where $m$ is the gradient?

Best Answer

Your work is correct, and it seems that your suggested answer is wrong. When the value of the slope is undefined, this points to the possibility that the tangent line is vertical. This is the case for your problem, so the tangent line should be x=2.

To see this more rigorously, you could consider $x$ as a function of $y$ and calculate $\frac{dx}{dy}$ at $(2,-3)$. You should get $0$. Then use this 'gradient' in the equation $m(y-y_1)=x-x_1$.

If the answer you've listed is correct, perhaps you have copied the problem down incorrectly.

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