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

Question

Elijah Allen · Accepted Answer

Please refer to the Explanation.

Suppose that the Normal # N : lx+my=1# meets the the Parabola # S : y^2=4ax# in the Parametric Point # P=P(at^2,2at)#. #P in N :. l(at^2)+m(2at)=1, i,e., lat^2+2amt=1......(ast)#. Now, the Slope of the tangent at #P" is "[dy/dx]_P#. #y^2=4ax rArr 2y*dy/dx=4a, or, dy/dx=(4a)/(2y)=(2a)/y#. #:. [dy/dx]_P=[(2a)/y]_(at^2,2at)=(2a)/(2at)=1/t#. #:.#The Slope of the Normal at #P" is "-t.# On the other hand, the slope of #N# is #-l/m#. #:. -t=-l/m, or, t=l/m#. Using this in #(ast), la(l/m)^2+2am(l/m)=1# #:. (al^3)/m^2+2al=1, or, al^3+2alm^2=m^3,# as desired!

Lincoln Anderson · Answer

undefined 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:

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

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

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

$$ l^2 + m^2 = -16a^2l^2m^2 $$

$$ l^2 + m^2 = -16a^2lm^2 $$

$$ l^2 + m^2 = -16a^2m $$

$$ m^2 = -l^2 - 16a^2m $$

$$ m^2 = -l^2 - 16a^2(2a) $$

$$ m^2 = -l^2 - 32a^3 $$

$$ al^3 + 2alm = m^2 $$