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: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$. The giv

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

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(C) Given the parabola $y^2 = 6x$. Differentiating with respect to $x$,we get: $2y \frac{dy}{dx} = 6 \Rightarrow \frac{dy}{dx} = \frac{3}{y}$. The given equation of the tangent is $5x - 2y + k = 0$,which can be written as $2y = 5x + k$,or $y = \frac{5}{2}x + \frac{k}{2}$. The slope of this tangent is $m = \frac{5}{2}$. Equating the slope of the tangent to the derivative at the point of contact: $\frac{3}{y} = \frac{5}{2} \Rightarrow y = \frac{6}{5}$. Substituting the value of $y$ into the parabola equation $y^2 = 6x$: $\left(\frac{6}{5}\right)^2 = 6x$ $\Rightarrow \frac{36}{25} = 6x$ $\Rightarrow x = \frac{6}{25}$. Thus,the point of contact is $\left(\frac{6}{25}, \frac{6}{5}\right)$.

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

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