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  • If the ratio of ordinates of two points on the parabola be <img alt="\mathrm{2: 1}"

If the ratio of ordinates of two points on the parabola \mathrm{y^2=4 x} be \mathrm{2: 1}, then the point of intersection of normals to parabola at these points describes the curve \mathrm{\left(\frac{x-2}{K}\right)^3=\frac{y^2}{36}}, where \mathrm{\mathrm{K}=.}

Option: 1

1


Option: 2

3


Option: 3

7


Option: 4

12


Answers (1)

best_answer

Let \mathrm{P\left(t_1{ }^2, 2 t_1\right)} and \mathrm{Q\left(t_2{ }^2, 2 t_2\right)} be two points on the parabola

now given that \mathrm{\frac{2 \mathrm{t}_2}{2 \mathrm{t}_1}=\frac{2}{1} \mathrm{t}_2=2 \mathrm{t}_1}                         ......(1)

Equations of normal at P and Q are \mathrm{y+t_1 x=2 t_1+t_1{ }^3}

and \mathrm{y+t_2 x=2 t_2+t_2{ }^3}                          ......(2)

On solving the equations of normals we get

\mathrm{ \begin{aligned} & X=2+t_1{ }^2+t_2{ }^2+t_1 t_2 \\\\ & X=2+t_1{ }_1+4 t_1{ }^2+2 t_1{ }_1\left[\begin{array}{ll} \text { Q } & t_2=2 t_1 \end{array}\right] \\\\ & X=2+7 t_1{ }^2\, \, \, \, \, ....(3)\\\\ & Y=-t_1 t_2\left(t_1+t_2\right)=-2 t_1{ }^2\left(3 t_1\right)=-6 t_1{ }^3 ....(4)\end{aligned} }

Eliminating \mathrm{t_1} in (3) and (4) we get

\mathrm{ \left(\frac{x-2}{7}\right)^3=\left(\frac{y}{6}\right)^2 }

Posted by

Shailly goel

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