Variable straight lines y=mx+c make intercept | vedclass

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(A) The equation of the parabola is $y^2 = 4ax$. The equation of the line is $y = mx + c$,which can be written as $\frac{y-mx}{c} = 1$.To find the com

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$(4a, 0)$

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(A) The equation of the parabola is $y^2 = 4ax$. The equation of the line is $y = mx + c$,which can be written as $\frac{y-mx}{c} = 1$.To find the combined equation of the lines joining the origin $(0,0)$ to the points of intersection of the line and the parabola,we homogenize the equation of the parabola using the line equation:$y^2 - 4ax \left( \frac{y-mx}{c} \right) = 0$$cy^2 - 4axy + 4amx^2 = 0$$4amx^2 - 4axy + cy^2 = 0$Since these lines subtend a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$4am + c = 0$Substituting $c = -4am$ into the line equation $y = mx + c$:$y = mx - 4am$$y = m(x - 4a)$This equation represents a family of lines passing through the fixed point $(4a, 0)$.Thus,the point of concurrence is $(4a, 0)$.

Variable straight lines $y=mx+c$ make intercepts on the curve $y^2-4ax=0$ which subtend a right angle at the origin. Then the point of concurrence of these lines $y=mx+c$ is

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