Conic-Sections Question 546
Question: A tangent to the parabola $ y^{2}=8x, $ which makes an angle of $ 45{}^\circ $ with the straight line $ y=3x+5 $ is
Options:
A) $ 2x-y+1=0 $
B) $ 2x+y+1=0 $
C) $ x-2y+8=0 $
D) Both &
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Answer:
Correct Answer: D
Solution:
[d] We know the tangent to the parabola $ y^{2}=4ax $ at $ (at^{2},2at) $ is $ ty=x+at^{2}. $ Here $ a=2 $ so, the tangent at $ (2t^{2},4t) $ to the parabola $ y^{2}=8x $ is $ ty=x+2t^{2} $ ?. (i) ?m? of (i) is $ \frac{1}{t}; $ (i) makes $ 45{}^\circ $ with $ y=3x+5 $ if $ \tan 45{}^\circ =| \frac{\frac{1}{t}-3}{1+\frac{1}{t}.3} |=| \frac{1-3t}{t+3} | $
$ \therefore 1=| \frac{1-3t}{t+3} |; $ Or $ \frac{1-3t}{t+3}=\pm 1; $ or $ 1-3t=t+3,-t-3 $
$ \therefore 4t=-2 $ Or $ 2t=4. $
$ \therefore t=-\frac{1}{2}or2 $
Putting in (i), the tangents have the equations
$ -\frac{1}{2}y=x+2.\frac{1}{4}i.e.,2x+y+1=0 $
and $ 2y=x+2.i.e.,x-2y+8=0 $
BETA
