The curve satisfying y'=x2+y2x2-y2,y(1)-0 has the slope at the point (1,0) of the curve equal to

Given y'=x2+y2x2−y2 and y(1)=0Slope = dydx⇒ dydx=x2+y2x2−y2⇒ (dydx)(1, 0)=1+01−0=1

The curve satisfying y'=x2+y2x2-y2,y(1)-0 has the slope at the point (1,0) of the curve equal to

Given y'=x2+y2x2−y2 and y(1)=0Slope = dydx⇒ dydx=x2+y2x2−y2⇒ (dydx)(1, 0)=1+01−0=1

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The curve satisfying $y' = \frac{x^{2} + y^{2}}{x^{2} - y^{2}}, y (1) - 0$ has the slope at the point (1,0) of the curve equal to

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$\frac{5}{3}$

$- \frac{5}{3}$

–1

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n y' = \frac{x^{2} + y^{2}}{x^{2} - y^{2}} a n d y (1) = 0 \\ S l o p e = \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x} = \frac{x^{2} + y^{2}}{x^{2} - y^{2}} \\ \Rightarrow {(\frac{d y}{d x})}_{(1, 0)} = \frac{1 + 0}{1 - 0} = 1 \end{array}$