# If lx+my=1 is the normal to the parabola y^2=4ax then prove that al^3+2alm=m^2 ?

Please refer to the Explanation.

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To prove ( al^3 + 2alm = m^2 ), we start with the given equation of the normal to the parabola ( y^2 = 4ax ), which is ( lx + my = 1 ).

Given that the normal intersects the parabola at ( (at^2, 2at) ), where ( t ) is the slope of the tangent at the point of intersection, we can find the slope of the normal using the relation ( mm' = -1 ), where ( m' ) is the slope of the normal.

From the equation of the normal, we have ( m' = -\frac{l}{m} ).

Given the point ( (at^2, 2at) ), we can find the slope of the tangent at this point by differentiating ( y^2 = 4ax ) with respect to ( x ), which gives ( \frac{dy}{dx} = \frac{1}{2a} ).

Thus, the slope of the tangent is ( \frac{1}{2a} ).

Since the product of the slopes of the tangent and the normal is -1, we have:

[ t \times m' = -1 ]

Substituting the slope of the tangent and the expression for ( m' ), we get:

[ t \times \left(-\frac{l}{m}\right) = -1 ]

[ -\frac{tl}{m} = -1 ]

[ tl = m ]

From the equation of the normal, ( lx + my = 1 ), substituting ( t = \frac{1}{2a} ) and ( tl = m ), we have:

[ l \times \frac{1}{2a} \times m = 1 ]

[ \frac{l \times m}{2a} = 1 ]

[ lm = 2a ]

Now, squaring the equation of the normal:

[ (lx + my)^2 = 1 ]

[ (lx)^2 + 2lmxy + (my)^2 = 1 ]

Substituting ( y^2 = 4ax ), we get:

[ (lx)^2 + 2lm(4ax) + (my)^2 = 1 ]

[ l^2x^2 + 8almx + m^2x^2 = 1 ]

[ (l^2 + m^2)x^2 + 8almx - 1 = 0 ]

For this quadratic equation to represent a pair of coincident lines, the discriminant must be zero:

[ (8alm)^2 - 4(l^2 + m^2)(-1) = 0 ]

[ 64a^2l^2m^2 + 4(l^2 + m^2) = 0 ]

[ 64a^2l^2m^2 + 4(l^2 + m^2) = 0 ]

[ 4(l^2 + m^2) = -64a^2l^2m^2 ]

[ l^2 + m^2 = -16a^2l^2m^2 ]

[ l^2 + m^2 = -16a^2lm^2 ]

[ l^2 + m^2 = -16a^2m ]

[ m^2 = -l^2 - 16a^2m ]

[ m^2 = -l^2 - 16a^2(2a) ]

[ m^2 = -l^2 - 32a^3 ]

[ al^3 + 2alm = m^2 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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