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If a variable circle passing through the points P, Q, R on the parabola $\mathrm{y^2=4 a x}$ , whose normals are concurrent at $\mathrm{(\alpha, \beta)}$ . The circle also passes through vertex of the parabola. Find out the centre of this circle.

Option: 1
$\mathrm{\left(\frac{\alpha+2 a}{2}, \frac{\beta}{4}\right)}$

Option: 2
$\mathrm{\left(\frac{\alpha-2 a}{2}, \beta / 4\right)}$

Option: 3
$\mathrm{(\alpha+2 a, \beta)}$

Option: 4
$\mathrm{(\alpha-2 a, \beta / 2)}$

Answers (1)

best_answer

Let the circle $\mathrm{x^2+y^2+2 g x+2 f y+c=0}$ …(i)

$\mathrm{y^2=4 a x}$ …(ii)

If we eliminate x from (i), (ii), we get

$\mathrm{\frac{y^4}{16 a^2}+y^2+\frac{2 g y^2}{4 a}+2 f y+c=0}$

$\mathrm{\text { or } \quad y^4+\left(16 a^2+8 a g\right) y^2+32 a^2 f y+16 a^2 c=0}$ …(iii)

Roots of (iii) will give ordinates of points of intersection of (i) and (ii). If $\mathrm{y_1, y_2, y_3, y_4}$ be its roots, then

$\mathrm{y1 + y2 + y3 + y4 = 0}$ …(1)

$\mathrm{y1 y2 + y1 y3 + }$ ……. $\mathrm{+ y3 y4 = 16a2 + 8ag }$ …(2)

$\mathrm{y1 y2 y3 + y1 y2 y4 + y1 y3 y4 + y2 y3 y4 =-32a2f}$ …(3)

$\mathrm{y1 y2 y3 y4 = 16a2c}$ …(4)

Since normals at P, Q, R are concurrent so $\mathrm{y_1+y_2+y_3=0}$

$\mathrm{y_4=0 \text { by (i) }}$

$\mathrm{x_4=0}$ , so circle always passes through (0, 0), which is vertex of the parabola.

Now if normals passing through (α, β) we have

$\mathrm{ y1 + y2 + y3 = 0}$ …(4)

$\mathrm{ y1 y2 + y2 y3 + y3 y4 = 4a(2a -\alpha )}$ ...(5)

$\mathrm{ y1 y2 y3 = 8a2\ss }$ …(6)

Put y4 = 0 in (2) and equate with (5),

$\mathrm{ \begin{aligned} & 16 a^2+8 a g=4 a(2 a-a) \\ & \Rightarrow \quad 2 g=-(a+2 a) \\ & \end{aligned} }$

Put y4 = 0 in (3) and equate with (6)

$\mathrm{ \begin{aligned} -32 a^2 f & =8 a^2 \beta \\ \Rightarrow \quad 2 f & =-\beta / 2 \end{aligned} }$

So equation of circle is $\mathrm{ x^2+y^2-(a+2 a) x-\frac{\beta}{2} y=0 }$

Posted by

manish painkra

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