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

Open in App

Solution

Let the parabola be $y^{2} = 4 a x$

Equation of normal at $(a t_{1}^{2}, 2 a t_{1})$ is

$y = - t_{1} x + 2 a t_{1} + a t_{1}^{3}$

$(a t_{2}^{2}, 2 a t_{2})$ also lies on the line

$∴ 2 a t_{2} = - t_{1} (a t_{2}^{2}) + 2 a t_{1} + a t_{1}^{3} 2 a t_{2} - 2 a t_{1} = a t_{1}^{3} - a t_{1} t_{2}^{2} - 2 a (t_{1} - t_{2}) = a t_{1} (t_{1}^{2} - t_{2}^{2}) - 2 (t_{1} - t_{2}) = t_{1} (t_{1} - t_{2}) (t_{1} + t_{2}) - \frac{2}{t_{1}} = t_{1} + t_{2} t_{2} = - \frac{2}{t_{1}} - t_{1}$

Hence proved.

Suggest Corrections

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.

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

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

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

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

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.

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

a t_{2}^{2}, 2 a t_{2}

t_{1} \neq t_{2}

Join BYJU'S Learning Program