The line that is normal to the curve $\displaystyle x^2 + xy - 2y^2 = 0 $ at $\displaystyle (4,4)$ intersects the curve at what other point?
I can not find an example of how to do this equation. Can someone help me out?
2026-03-31 10:08:41.1774951721
The normal line intersects a curve at two points. What is the other point?
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$$x^2 + xy - 2y^2 = 0$$ $$2x+(y+xy^\prime)-4yy^\prime=0$$ $$2x+y=4yy^\prime-xy^\prime$$ $$2x+y=y^\prime(4y-x)$$ $$\dfrac{2x+y}{4y-x}=y^\prime$$
There's the expression differentiated. I'm assuming that's what you had a problem with. The next step would be evaluating this expression a (4,4). The find the gradient of the normal line ($m_{normal}\cdot m=-1$).
Once you have the gradient of the normal line (denoted 'm'), solve for c in y=mx+c at point (4,4). This will yield an equation of a line and the equation this equation of line to the original expression and solve for x and y.
I leave the details of the solution for you to work out. This is a strategy as to how to go about solving it.
Good luck!