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  • A point  moves such that sum of the slopes of the normals drawn from it to the hyperbola <img alt

A point \mathrm{P} moves such that sum of the slopes of the normals drawn from it to the hyperbola \mathrm{x y=4} is equal to the sum of ordinates of feet of normals. Prove that the locus of \mathrm{P} is a parabola. Find the least distance of this parabola from the circle \mathrm{x^2+y^2-24 x+128=0}

Option: 1

\sqrt{80}-4


Option: 2

\sqrt{80}-3


Option: 3

\sqrt{70}-4


Option: 4

\sqrt{70}-3


Answers (1)

best_answer

Any point on the hyperbola \mathrm{x y=4 is (2 t, 2 / t) }
Now normal at \mathrm{(2 t, 2 / t) is y-2 / t=t^2(x-2 t) (its\, slope\, is\, t^2 ) }
If the normal passes through \mathrm{\mathrm{P}(\mathrm{h}, \mathrm{k}) then \mathrm{k}-2 / \mathrm{t}=\mathrm{t}^2(\mathrm{~h}-2 \mathrm{t}) }
\mathrm{\Rightarrow \quad 2 \mathrm{t}^4-\mathrm{ht}^3+\mathrm{tk}-2=0..............(1) }
roots of (1) given parameters of feet of normals passing through (h, k).
Let roots be \mathrm{t_1, t_2, t_3 \, and\, t_4, } then
\mathrm{\begin{aligned} & t_1+t_2+t_3+t_4=h / 2 ........................(2)\\ & t_1 t_2+t_1 t_3+t_1 t_4+t_2 t_4+t_2 t_3+t_3 t_4=0.....................(3) \\ & t_1 t_2 t_3+t_1 t_2 t_4+t_1 t_3 t_4+t_2 t_3 t_4=-k / 2................(4)\\ & t_1 t_2 t_3 t_4=-1.............................(5) \end{aligned} }
\mathrm{\begin{aligned} & \frac{\text { (4) }}{\text { (5) } \Rightarrow} \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}=\frac{k}{2} \\ & \Rightarrow \quad \mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3+\mathrm{y}_4=\mathrm{k} \\ & \end{aligned} }
From (2) and (3);
\mathrm{t_1^2+t_2^2+t_3^2+t_4^2=\frac{h^2}{4} }
Given that \mathrm{\frac{h^2}{4}=k \Rightarrow locus \, \, of\, \, (h, k) is x^2=4 y. } Which is a parabola.
Now distance between this parabola and given circle will be along the common normal.
Normal at \mathrm{\left(2 \mathrm{~m}, \mathrm{~m}^2\right). } on the parabola \mathrm{\mathrm{x}^2=4 \mathrm{y} \, \, is \, \, \mathrm{x}+\mathrm{my}=2 \mathrm{~m}+\mathrm{m}^3.. } If it is common normal to parabola and circle, then it passes through the centre (12,0) of circle,
\mathrm{\begin{array}{rlrl} & \text { so } & 12=2 \mathrm{~m}+\mathrm{m}^3 \\ \Rightarrow & \mathrm{m}=2 \end{array} }
So least distance \mathrm{=\sqrt{(4-12)^2+(4-0)^2}-4=(\sqrt{80}-4) }units.

Posted by

Kuldeep Maurya

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