question 6 of 15, step 1 of 1 correct find all points on the curve 5xy² - 10x²y = 5 where it has a…

Explanation:

Step1: Differentiate implicitly

Differentiate $5xy^{2}-10x^{2}y = 5$ with respect to $x$. Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$, we have: $5(y^{2}+2xyy^\prime)-10(2xy + x^{2}y^\prime)=0$. Expand: $5y^{2}+10xyy^\prime-20xy - 10x^{2}y^\prime=0$.

Step2: Solve for $y^\prime$

Group the terms with $y^\prime$: $10xyy^\prime-10x^{2}y^\prime=20xy - 5y^{2}$. Factor out $y^\prime$: $y^\prime(10xy - 10x^{2})=20xy - 5y^{2}$. So, $y^\prime=\frac{20xy - 5y^{2}}{10xy - 10x^{2}}=\frac{4xy - y^{2}}{2xy - 2x^{2}}$.

Step3: Set $y^\prime = 0$

For a horizontal tangent line, $y^\prime = 0$. So, $4xy - y^{2}=0$. Factor out $y$: $y(4x - y)=0$. Case 1: $y = 0$ Substitute $y = 0$ into the original equation $5xy^{2}-10x^{2}y = 5$. We get $0=5$, which is a contradiction. Case 2: $y = 4x$ Substitute $y = 4x$ into the original equation $5x(4x)^{2}-10x^{2}(4x)=5$. $5x\times16x^{2}-40x^{3}=5$. $80x^{3}-40x^{3}=5$. $40x^{3}=5$. $x^{3}=\frac{1}{8}$. $x=\frac{1}{2}$. When $x = \frac{1}{2}$, $y=4\times\frac{1}{2}=2$.

Answer:

$(\frac{1}{2},2)$