Question

verify that the given point is on the curve and find the line tangent to the curve at the give point.
$x^{2}+xy - y^{2}=1$, $(2,3)$

Explanation:

Step1: Verify the point is on the curve

Substitute \(x = 2\) and \(y = 3\) into the equation \(x^{2}+xy - y^{2}\).

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Since the result is \(1\) (equal to the right - hand side of the equation), the point \((2,3)\) is on the curve.

Step2: Differentiate the equation implicitly

Differentiate \(x^{2}+xy - y^{2}=1\) with respect to \(x\).
Using the sum rule \((u + v+w)^\prime=u^\prime + v^\prime+w^\prime\), where \(u = x^{2}\), \(v = xy\), \(w=-y^{2}\).

• For \(u=x^{2}\), \(u^\prime = 2x\).
• For \(v = xy\), use the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (here \(u = x\), \(v = y\)), so \(v^\prime=y + x\frac{dy}{dx}\).
• For \(w=-y^{2}\), use the chain rule \(w^\prime=-2y\frac{dy}{dx}\).

The derivative of the left - hand side is \(2x+y + x\frac{dy}{dx}-2y\frac{dy}{dx}\), and the derivative of the right - hand side is \(0\). So \(2x+y + x\frac{dy}{dx}-2y\frac{dy}{dx}=0\).

Step3: Solve for \(\frac{dy}{dx}\)

Rearrange the equation \(2x+y + x\frac{dy}{dx}-2y\frac{dy}{dx}=0\) to isolate \(\frac{dy}{dx}\).

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Step4: Find the slope of the tangent at \((2,3)\)

Substitute \(x = 2\) and \(y = 3\) into \(\frac{dy}{dx}=\frac{-2x - y}{x - 2y}\).

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Step5: Use the point - slope form to find the tangent line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(2,3)\) and \(m=\frac{7}{4}\).

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Answer:

The point \((2,3)\) is on the curve. The equation of the tangent line is \(7x - 4y-2 = 0\)