If 5x - 2y + k = 0 is a tangent to the parabo | vedclass

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(C) Given the parabola $y^2 = 6x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6$,which implies $\frac{dy}{dx} = \frac{3}{y}$. Give

Vedclass · Accepted Answer

$(\frac{6}{25}, \frac{6}{5})$

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(C) Given the parabola $y^2 = 6x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6$,which implies $\frac{dy}{dx} = \frac{3}{y}$. Given the tangent line $5x - 2y + k = 0$,we can rewrite it as $y = \frac{5}{2}x + \frac{k}{2}$. The slope of this line is $m = \frac{5}{2}$. At the point of contact,the slope of the tangent to the parabola must equal the slope of the line: $\frac{3}{y} = \frac{5}{2}$. Solving for $y$,we get $y = \frac{6}{5}$. Substituting $y = \frac{6}{5}$ into the parabola equation $y^2 = 6x$: $(\frac{6}{5})^2 = 6x$ $\frac{36}{25} = 6x$ $x = \frac{36}{25 	imes 6} = \frac{6}{25}$. Thus,the point of contact is $(\frac{6}{25}, \frac{6}{5})$.

If $5x - 2y + k = 0$ is a tangent to the parabola $y^2 = 6x$,then their point of contact is

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