JEE MainMathematicsToughParabola

Let normal's at two points on the parabola y^2 =4x intersects
on the parabola. If the line joining these two points passes through the
fixed point (α, β) then find |α+β|.
ekuk ijoy; y^2 =4x ij nks fcUnqvks ij ij çfrPNsn djrs
bu nks fcUnqvksa ls xqtjus okyh,d fLFkj fcUnq (α, β)

Show answer and solution

The normal at P(t_1) on y^2 =4x is

y - 2t_{1} = - t_{1}\left( x - t_{1}^{2} \right)

..........(i)
The normal at

Q\left( t_{2} \right)\ is\ y - 2t_{2} = - t_{2}\left( x - t_{2}^{2} \right)

Suppose (1) and (2) cut at (t^2, 2t) on the parabola then

t = - \frac{2}{t_{1}} - t_{1} = - \frac{2}{t_{2}} - t_{2}

\Rightarrow \frac{2\left( t_{1} - t_{2} \right)}{t_{2}t_{2}} = t_{1} - t_{2} \Rightarrow t_{1}t_{2} = 2

the equation of the chord PQ is

y\left( t_{1} + t_{2} \right) = 2x + 2t_{1}t_{2}

\Rightarrow 2x - y\left( t_{1} + t_{2} \right) + 4 = 0

Which obviously passes through the intersection of 2x+4=0 and y=0
i.e. (-2,0) for all values of t_1 and t_2
Hindi. ijoy; y^2 =4x ij P(t_1) lehdj.k

y - 2t_{1} = - t_{1}\left( x - t_{1}^{2} \right)

..........(i)

Q\left( t_{2} \right)\

lehdj.k

y - 2t_{2} = - t_{2}\left( x - t_{2}^{2} \right)

t = - \frac{2}{t_{1}} - t_{1} = - \frac{2}{t_{2}} - t_{2}

\Rightarrow \frac{2\left( t_{1} - t_{2} \right)}{t_{2}t_{2}} = t_{1} - t_{2} \Rightarrow t_{1}t_{2} = 2

thok PQ dk lehdj.k

y\left( t_{1} + t_{2} \right) = 2x + 2t_{1}t_{2}

\Rightarrow 2x - y\left( t_{1} + t_{2} \right) + 4 = 0

vFkkZr t_1vkSj t_2ds lHkh ekuks

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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