At which point does the line x = my + frac{a} | vedclass

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(D) Step $1$: Express $y$ in terms of $x$ from the line equation.Given $x = my + \frac{a}{m}$,we have $my = x - \frac{a}{m}$,so $y = \frac{x}{m} - \fr

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$\left( \frac{2a}{m}, \frac{a}{m^2} \right)$

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(D) Step $1$: Express $y$ in terms of $x$ from the line equation.Given $x = my + \frac{a}{m}$,we have $my = x - \frac{a}{m}$,so $y = \frac{x}{m} - \frac{a}{m^2}$.Step $2$: Substitute $y$ into the parabola equation $x^2 = 4ay$.$x^2 = 4a \left( \frac{x}{m} - \frac{a}{m^2} \right)$$x^2 = \frac{4ax}{m} - \frac{4a^2}{m^2}$$m^2x^2 - 4amx + 4a^2 = 0$Step $3$: Solve the quadratic equation for $x$.$(mx - 2a)^2 = 0$$mx = 2a \implies x = \frac{2a}{m}$.Step $4$: Find the corresponding $y$ coordinate.$y = \frac{x}{m} - \frac{a}{m^2} = \frac{2a}{m^2} - \frac{a}{m^2} = \frac{a}{m^2}$.Thus,the point of contact is $\left( \frac{2a}{m}, \frac{a}{m^2} \right)$.

At which point does the line $x = my + \frac{a}{m}$ touch the parabola $x^2 = 4ay$?

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