Question

2 mark for review which of the following is true about the curve $x^2 - xy + y^2 = 3$ at the point $(2, 1)$? a $\frac{dy}{dx}$ exists at $(2, 1)$, but there is no tangent line at that point. b $\frac{dy}{dx}$ exists at $(2, 1)$, and the tangent line at that point is horizontal. c $\frac{dy}{dx}$ exists at $(2, 1)$, and the tangent line at that point is neither horizontal nor vertical. d $\frac{dy}{dx}$ does not exist at $(2, 1)$, and the tangent line at that point is vertical. e $\frac{dy}{dx}$ does not exist at $(2, 1)$, and the tangent line at that point is horizontal.

Explanation:

Step1: Implicit Differentiation

Differentiate \(x^2 - xy + y^2 = 3\) with respect to \(x\).
Using product rule for \(-xy\): \(\frac{d}{dx}(-xy)=-y - x\frac{dy}{dx}\), \(\frac{d}{dx}(x^2)=2x\), \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\), \(\frac{d}{dx}(3)=0\).
So, \(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Rearrange terms: \(-x\frac{dy}{dx}+2y\frac{dy}{dx}=y - 2x\)
Factor \(\frac{dy}{dx}\): \(\frac{dy}{dx}(-x + 2y)=y - 2x\)
Thus, \(\frac{dy}{dx}=\frac{y - 2x}{-x + 2y}=\frac{2x - y}{x - 2y}\).

Step3: Evaluate \(\frac{dy}{dx}\) at \((2,1)\)

Substitute \(x = 2\), \(y = 1\) into \(\frac{dy}{dx}\): \(\frac{2(2)-1}{2 - 2(1)}=\frac{4 - 1}{2 - 2}=\frac{3}{0}\)? Wait, no, wait: Wait, \(\frac{2x - y}{x - 2y}\) at \((2,1)\): numerator \(2*2 - 1 = 3\), denominator \(2 - 2*1 = 0\)? Wait, no, I made a mistake in step 2. Wait, original differentiation: \(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\) → \((2y - x)\frac{dy}{dx}=y - 2x\) → \(\frac{dy}{dx}=\frac{y - 2x}{2y - x}\). Ah, that's the error. So correct \(\frac{dy}{dx}=\frac{y - 2x}{2y - x}\).

Now substitute \(x = 2\), \(y = 1\): \(\frac{1 - 4}{2 - 2}=\frac{-3}{0}\)? No, wait, no: \(2y - x\) when \(x=2,y=1\) is \(2*1 - 2 = 0\)? Wait, no, \(2y - x = 2(1)-2 = 0\), \(y - 2x = 1 - 4 = -3\). Wait, that can't be. Wait, let's redo differentiation:

\(\frac{d}{dx}(x^2 - xy + y^2)=\frac{d}{dx}(3)\)

\(2x - (y + x\frac{dy}{dx}) + 2y\frac{dy}{dx}=0\)

\(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\)

\(2x - y + \frac{dy}{dx}(2y - x)=0\)

\(\frac{dy}{dx}(2y - x)=y - 2x\)

\(\frac{dy}{dx}=\frac{y - 2x}{2y - x}\). Now at \((2,1)\): \(y - 2x = 1 - 4 = -3\), \(2y - x = 2 - 2 = 0\). Wait, that would mean \(\frac{dy}{dx}\) is undefined? But that contradicts. Wait, no, maybe I messed up the sign. Wait, \(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\) → \( - x\frac{dy}{dx}+2y\frac{dy}{dx}=y - 2x\) → \(\frac{dy}{dx}(2y - x)=y - 2x\) → \(\frac{dy}{dx}=\frac{y - 2x}{2y - x}\). So at \((2,1)\), denominator is \(0\), numerator is \(-3\). So \(\frac{dy}{dx}\) is undefined? But that's not matching. Wait, maybe the original curve: check if \((2,1)\) is on the curve: \(2^2 - 2*1 + 1^2 = 4 - 2 + 1 = 3\), yes. Wait, maybe my differentiation is wrong. Let's try again.

Differentiate term by term:

- \(\frac{d}{dx}(x^2) = 2x\)
- \(\frac{d}{dx}(-xy) = -y - x\frac{dy}{dx}\) (product rule: derivative of x is 1, times y, plus x times derivative of y)
- \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\) (chain rule)
- \(\frac{d}{dx}(3) = 0\)

So combining: \(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\)

Now, collect \(\frac{dy}{dx}\) terms: \(\frac{dy}{dx}(2y - x) = y - 2x\)

Thus, \(\frac{dy}{dx}=\frac{y - 2x}{2y - x}\). Now, at (2,1):

Numerator: \(1 - 4 = -3\)

Denominator: \(2*1 - 2 = 0\)

So \(\frac{dy}{dx}\) is undefined, which would mean vertical tangent? But the options: D says \(\frac{dy}{dx}\) does not exist, tangent is vertical. Wait, but earlier mistake. Wait, maybe I flipped numerator and denominator. Wait, let's rearrange again:

\(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\)

\(2x - y = x\frac{dy}{dx}-2y\frac{dy}{dx}\)

\(2x - y = \frac{dy}{dx}(x - 2y)\)

Ah! Here's the mistake. I had a sign error. So \(x\frac{dy}{dx}-2y\frac{dy}{dx}=2x - y\) → \(\frac{dy}{dx}(x - 2y)=2x - y\) → \(\frac{dy}{dx}=\frac{2x - y}{x - 2y}\). Yes! That's the correct rearrangement. So earlier step 2 was wrong. So:

From \(2x - y - x\frac{dy}{dx}+2y\frac{dy}{dx}=0\)

Bring \(-x\frac{dy}{dx}+2y\frac{dy}{dx}\) to right and \(2x - y\) to left? No, move terms with \(\frac{dy}{dx}\) to one side:

\(-x\frac{dy}{dx}+2y\frac{dy}{dx}=y - 2x\) → no, wait, add \(x\frac{dy}{dx}\) and subtract \(y\) from both sides:

\(2x - y = x\frac{dy}{dx}-2y\frac{dy}{dx}\) → \(2x - y = \frac{dy}{dx}(x - 2y)\) → \(\frac{dy}{dx}=\frac{2x - y}{x - 2y}\). Yes, that's correct. So my earlier sign was wrong. So now, at (2,1):

Numerator: \(2*2 - 1 = 3\)

Denominator: \(2 - 2*1 = 0\). Wait, denominator is 0? No, \(x - 2y = 2 - 2*1 = 0\), numerator is 3. So \(\frac{dy}{dx}\) is undefined, meaning vertical tangent? But option D says \(\frac{dy}{dx}\) does not exist, tangent is vertical. But wait, the original selected option was C. Wait, maybe I made a mistake. Wait, let's check the curve \(x^2 - xy + y^2 = 3\) at (2,1). Let's plot mentally: it's a conic section. Let's check the derivative again.

Wait, \(\frac{dy}{dx}=\frac{2x - y}{x - 2y}\). At (2,1): (4 - 1)/(2 - 2) = 3/0, which is undefined. So \(\frac{dy}{dx}\) does not exist, and the tangent is vertical. But the options: D is " \(\frac{dy}{dx}\) does not exist at (2, 1), and the tangent line at that point is vertical." But in the image, D was crossed out. Wait, maybe the original problem's options were misread. Wait, let's re-express the differentiation correctly.

Wait, let's do implicit differentiation again carefully:

Given \(F(x,y)=x^2 - xy + y^2 - 3 = 0\)

Using implicit differentiation formula: \(\frac{dy}{dx}=-\frac{F_x}{F_y}\)

Compute \(F_x = 2x - y\)

Compute \(F_y = -x + 2y\)

Thus, \(\frac{dy}{dx}=-\frac{2x - y}{-x + 2y}=\frac{2x - y}{x - 2y}\). Yes, that's correct. So \(F_x = 2x - y\), \(F_y = -x + 2y\). Then \(\frac{dy}{dx}=-\frac{F_x}{F_y}=\frac{2x - y}{x - 2y}\).

Now, at (2,1): \(F_x = 4 - 1 = 3\), \(F_y = -2 + 2 = 0\). So \(\frac{dy}{dx}=-\frac{3}{0}\), which is undefined. So the derivative does not exist, and the tangent line is vertical (since the derivative is undefined, tangent is vertical). But the option D is " \(\frac{dy}{dx}\) does not exist at (2, 1), and the tangent line at that point is vertical." But in the image, D was crossed out. Wait, maybe there's a mistake in my calculation. Wait, let's check the curve at (2,1): plug x=2, y=1: 4 - 2 + 1 = 3, which matches. Now, let's see the slope: if \(\frac{dy}{dx}\) is undefined, tangent is vertical. So option D should be correct. But the original selected option was C. Wait, maybe I messed up \(F_y\). Wait, \(F_y\) is derivative of \(F\) with respect to y: derivative of \(x^2\) with respect to y is 0, derivative of \(-xy\) with respect to y is -x, derivative of \(y^2\) with respect to y is 2y, derivative of -3 is 0. So \(F_y = -x + 2y\). Correct. Then \(\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{2x - y}{-x + 2y}=\frac{2x - y}{x - 2y}\). At (2,1): (4 - 1)/(2 - 2) = 3/0, undefined. So tangent is vertical, so \(\frac{dy}{dx}\) does not exist. So option D is correct. But the image shows D crossed out. Maybe the original problem had a typo, or I made a mistake. Wait, alternatively, maybe the curve is \(x^2 - xy + y^2 = 3\), let's check the derivative at (1,2) instead? No, the point is (2,1). Wait, maybe I flipped x and y. Wait, no, the point is (2,1). Wait, let's compute the slope again. Wait, \(\frac{dy}{dx}=\frac{2x - y}{x - 2y}\). At (2,1): (4 - 1)/(2 - 2) = 3/0, undefined. So tangent is vertical, so \(\frac{dy}{dx}\) does not exist. So option D: " \(\frac{dy}{dx}\) does not exist at (2, 1), and the tangent line at that point is vertical." is correct. But in the image, D was crossed out, and C was selected. Maybe there's a mistake in my differentiation. Wait, let's try another approach. Let's solve for y in terms of x, but it's a quadratic in y: \(y^2 - xy + x^2 - 3 = 0\). Using quadratic formula: \(y = \frac{x \pm \sqrt{x^2 - 4(x^2 - 3)}}{2} = \frac{x \pm \sqrt{12 - 3x^2}}{2}\). At x=2, \(12 - 12 = 0\), so \(y = \frac{2 \pm 0}{2} = 1\). So the curve has a vertical tangent at (2,1) because the quadratic in y has a repeated root (discriminant zero) at x=2, meaning the curve is tangent to the vertical line x=2 at (2,1). Thus, the derivative \(\frac{dy}{dx}\) does not exist (vertical tangent), so option D is correct. But the original selected option was C, which is wrong. Wait, maybe the problem was \(x^2 - xy + y^2 = 3\) at (1,2)? Let's check (1,2): \(\frac{dy}{dx}=\frac{2*1 - 2}{1 - 4}=\frac{0}{-3}=0\), which would be horizontal tangent. But the point is (2,1). So there must be a mistake in the initial differentiation. Wait, no, the correct differentiation gives \(\frac{dy}{dx}\) undefined at (2,1), so tangent is vertical, so option D is correct.

Answer:

D. \(\frac{dy}{dx}\) does not exist at \((2, 1)\), and the tangent line at that point is vertical.