Conic Sections Question 115
Question: If The tangent to the parabola $ y^{2}=ax $ makes an angle of 45o with x-axis, then the point of contact is
[RPET 1985, 90, 2003]
Options:
A) $ ( \frac{a}{2},\ \frac{a}{2} ) $
B) $ ( \frac{a}{4},\ \frac{a}{4} ) $
C) $ ( \frac{a}{2},\ \frac{a}{4} ) $
D) $ ( \frac{a}{4},\ \frac{a}{2} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
Parabola is $ y^{2}=ax $ i.e., $ y^{2}=4( \frac{a}{4} )x $ …..(i) $ \because $ Let point of contact is $ (x_1,y_1) $
Equation of tangent is $ y-y_1=\frac{2( \frac{a}{4} )}{y_1}(x-x_1) $
therefore $ y=\frac{a}{2y_1}(x)-\frac{ax_1}{2y_1}+y_1 $
Here, $ m=\frac{a}{2y_1}=\tan 45^{o} $
$ \Rightarrow $
$ \frac{a}{2y_1}=1 $
therefore $ y_1=\frac{a}{2} $
From (i), $ x_1=\frac{a}{4} $ . Point is $ ( \frac{a}{4},\frac{a}{2} ) $ .
BETA
