If the segment intercepted by the parabola y^ | vedclass

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(A) The equation of the line is $lx + my + n = 0$,which can be written as $\frac{lx + my}{-n} = 1$.The equation of the parabola is $y^2 = 4ax$,which c

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$4al + n = 0$

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(A) The equation of the line is $lx + my + n = 0$,which can be written as $\frac{lx + my}{-n} = 1$.The equation of the parabola is $y^2 = 4ax$,which can be written as $y^2 = 4ax(1)$.Substituting the value of $1$ from the line equation into the parabola equation,we get the homogeneous equation of the pair of lines passing through the vertex:$y^2 = 4ax \left( \frac{lx + my}{-n} \right)$$-ny^2 = 4alx^2 + 4amxy$$4alx^2 + 4amxy + ny^2 = 0$.Since these lines subtend a right angle at the vertex,the sum of the coefficients of $x^2$ and $y^2$ must be zero.Therefore,$4al + n = 0$.

If the segment intercepted by the parabola $y^2 = 4ax$ with the line $lx + my + n = 0$ subtends a right angle at the vertex,then

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