Find the relation between $t_{1} a n d t_{2}$ if normals at $(a t_{1}^{2}, 2 a t_{1}) a n d (a t_{2}^{2}, 2 a t_{2})$ meet on the parabola.

$t_{1} t_{2} = 2$

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$t_{1} + t_{2} = 2$

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$t_{1} t_{2} = 4$

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$t_{1} t_{2} = 1$

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Solution

The correct option is A
$t_{1} t_{2} = 2$

Suppose that two lines have the equations

y=ax+c and y=bx+d

Hence, the point of intersection is

$P (\frac{d - c}{a - b}, a \frac{d - c}{a - b} + c) = P (\frac{d - c}{a - b}, \frac{a d - b c}{a - b})$

Here normal are $y = - t_{1} x + a t_{1}^{3} + 2 a t_{1} a n d y = - t 2 x + a t_{2}^{3} + 2 a t_{2}$

Therefore $a = - t_{1}, b = - t_{2}, c = a t_{1}^{3} + 2 a t_{1}, d = a t_{2}^{3} + 2 a t_{2}$

And x and y coordinate of P lie on $y^{2} = 4 a x$ which gives us on simplification $t_{1} t_{2} = 2$

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Similar questions

Find the relation between $t_{1} a n d t_{2}$ if normals at $(a t_{1}^{2}, 2 a t_{1}) a n d (a t_{2}^{2}, 2 a t_{2})$ meet on the parabola.

Q. The normal at the point

(a t_{1}^{2}, 2 a t_{1})

meets the parabola again in the point

(a t_{2}^{2}, 2 a t_{2})

; prove that

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

Q. The normal at the point

(a t_{1}^{2}, 2 a t_{1})

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(a t_{2}^{2}, 2 a t_{2})

, then

If the chord joining the points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ of the parabola $y^{2} = 4 a x$ passes through the focus of the parabola, then

Q. If the normal at

t_{1}

to the hyperbola

x y = c^{2}

meets the hyperbola at

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. then find the relation between

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Find the relation between t1 and t2 if normals at (at21,2at1) and (at22,2at2) meet on the parabola.

Find the relation between $t_{1} a n d t_{2}$ if normals at $(a t_{1}^{2}, 2 a t_{1}) a n d (a t_{2}^{2}, 2 a t_{2})$ meet on the parabola.