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Exercise 7.2 · Q4

Q.Find the points on the curve y2−4xy=x2+5y^2-4xy=x^2+5 for which the tangent is horizontal.

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Differentiate implicitly, set the numerator of y′y' to zero (horizontal tangent), then use that relation together with the original equation to pin down the points.

Step 1. Differentiate y2−4xy=x2+5y^2-4xy=x^2+5 implicitly w.r.t. xx.

2y y′−4(y+x y′)=2x ⇒ y′(2y−4x)=2x+4y ⇒ y′=2x+4y2y−4x=x+2yy−2x2y\,y'-4(y+x\,y')=2x\ \Rightarrow\ y'(2y-4x)=2x+4y\ \Rightarrow\ y'=\dfrac{2x+4y}{2y-4x}=\dfrac{x+2y}{y-2x}.

Step 2. Horizontal tangent ⇒y′=0⇒\Rightarrow y'=0 \Rightarrow numerator =0=0.

x+2y=0⇒x=−2yx+2y=0\Rightarrow x=-2y.

Step 3. Substitute into the original curve equation.

y2−4(−2y)(y)=(−2y)2+5 ⇒ y2+8y2=4y2+5 ⇒ 5y2=5 ⇒ y2=1 ⇒ y=±1y^2-4(-2y)(y)=(-2y)^2+5\ \Rightarrow\ y^2+8y^2=4y^2+5\ \Rightarrow\ 5y^2=5\ \Rightarrow\ y^2=1\ \Rightarrow\ y=\pm1.

Step 4. Find xx for each yy. …

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