Let RS be the diameter of the circle , where is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q para

Question

Let RS be the diameter of the circle $x^{2} + y^{2} = 1$ , where $S$ is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s). [2016]

Smart Achievers · Accepted Answer

(1, 3)

Given: A circle : $x^{2} + y^{2} = 1$

Let coordinates of $P = (\cos θ, \sin θ)$

$∴$ $Equation of tangent at P (\cos θ, \sin θ) is$

$x \cos θ + y \sin θ = 1 ...(i)$

$Equation of normal at P is$ $y = x \tan θ ...(ii)$

$Now, equation of tangent at S is x = 1 ...(iii)$

$On solving (i) and (iii), we get the coordinates of Q as$

$(1, \frac{1 - \cos θ}{\sin θ}) = (1, \tan \frac{θ}{2})$

$∴$ $Equation of line through Q and parallel to R S is$ $y = \tan \frac{θ}{2} ...(iv)$

$Intersection point E of normal (ii) and line (iv) can be find out by solving (ii) and (iv).$

$Now from (ii) and (iv),$

$\tan \frac{θ}{2} = x \tan θ \Rightarrow x = \frac{1 - \tan^{2} \frac{θ}{2}}{2}$

$∴$ $Locus of E is$ $x = \frac{1 - y^{2}}{2} \Rightarrow y^{2} = 1 - 2 x$

$It is satisfied by the points$ $(\frac{1}{3}, \frac{1}{\sqrt{3}}) and (\frac{1}{3}, - \frac{1}{\sqrt{3}})$

Solution

1 $(\frac{1}{3}, \frac{1}{\sqrt{3}})$

2 $(\frac{1}{4}, \frac{1}{2})$

3 $(\frac{1}{3}, - \frac{1}{\sqrt{3}})$

4 $(\frac{1}{4}, - \frac{1}{2})$

Want Us to Email you About Special Offers & Updates?

Solution

1 (13,13)

2 (14,12)

3 (13,-13)

4 (14,-12)

Want Us to Email you About Special Offers & Updates?

1 $(\frac{1}{3}, \frac{1}{\sqrt{3}})$

2 $(\frac{1}{4}, \frac{1}{2})$

3 $(\frac{1}{3}, - \frac{1}{\sqrt{3}})$

4 $(\frac{1}{4}, - \frac{1}{2})$