Conic-Sections Question 573
Question: The equation of the parabola whose focus is (0, 0) and the tangent at the vertex is $ 8x-y+1=0 $ is
Options:
A) $ x^{2}+y^{2}+2xy-4x+4y-4=0 $
B) $ x^{2}-4x+4y-4=0 $
C) $ y^{2}-4x+4y-4=0 $
D) $ 2x^{2}+2y^{2}-4xy-x+y-4=0 $
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Answer:
Correct Answer: A
Solution:
[a] The length of the perpendicular drawn from the given focus upon the given line $ x-y+1=0 $ is $ \frac{0-0+1}{\sqrt{{{(1)}^{2}}+{{(-1)}^{2}}}}=\frac{1}{\sqrt{2}}. $ The directrix is parallel to the tangent at the vertex. So, the equation of the directrix is $ x-y+\lambda =0, $ Where $ \lambda $ is a constant to be determine? But the distance between the focus and the directrix $ =2\times $ (the distance between the focus and the tangent at the vertex) $ =2\times \frac{1}{\sqrt{2}}=\sqrt{2}. $ Hence $ \frac{0-0+\lambda }{\sqrt{{{(1)}^{2}}+{{(-1)}^{2}}}}=\sqrt{2}. $
$ \therefore \lambda =2. $ [ $ \lambda $ Must be positive see figure]
$ \therefore $ The directrix is the line $ x-y+2=0 $ . Let (x, y) be a moving point on the parabola. By The focus-directrix property of the parabola, its equation is $ {{(x-0)}^{2}}+{{(y-0)}^{2}}={{( \pm \frac{x-y+2}{\sqrt{2}} )}^{2}} $ Or $ x^{2}+y^{2}+2xy-4x+4y-4=0 $
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