Conic Sections Question 529
Question: The tangent and normal at the point $ P(at^{2},2at) $ to the parabola $ y^{2}=4ax $ meet the x-axis at T and G, respectively, Then the angle at which the tangent at p to the parabola is inclined to the tangent at p to the circle through P, T and G is
Options:
A) $ {{\tan }^{-1}}(t^{2}) $
B) $ {{\cot }^{-1}}(t^{2}) $
C) $ {{\tan }^{-1}}(t) $
D) $ {{\cot }^{-1}}(t) $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Tangent and normal at $ P(at^{2},2at) $ to the parabola $ y^{2}=4ax $ are, respectively, $ ty=x+at^{2} $
$ andy=-tx+2at+at^{3} $ Equations (i) and (ii) meet the x-axis where y=0. From (i), $ x=-at^{2} $
Therefore, T is $ (-at^{2},0) $ .
Form (ii), $ tx=2at+at^{3} $ . Therefore, G is $ (2a+at^{3},0). $ Midpoint of TG= $ =( \frac{2a+at^{2}-at^{2}}{2} )\equiv 0(a,0) $ Since $ \angle TPG=90{}^\circ $ , the center of the circle through PTG is (a, 0).
If $ \theta $ is the angle between tangents at P to the parabola and circle through P, T and G then $ (90{}^\circ -\theta ) $ is the angle between PT and OP.
Slope of PT= $ \frac{2at}{2at^{2}}=\frac{1}{t} $
Slope of OP $ =\frac{2at}{a(t^{2}-1)}=\frac{2t}{r^{2}-1} $
$ \therefore \tan (90{}^\circ -\theta )=| \frac{\frac{1}{t}-\frac{2t}{r^{2}-1}}{1+\frac{1}{t}( \frac{2t}{r^{2}-1} )} |=\frac{1}{t} $
$ \therefore \cot \theta =\frac{1}{t} $ Or $ \tan \theta =t $ Or $ \theta ={{\tan }^{-1}}(t) $
BETA
