Conic Sections Question 323
Question: The equation of the parabola whose focus is the point (0, 0) and the tangent at the vertex is $ x-y+1=0 $ is
[Orissa JEE 2002]
Options:
A) $ x^{2}+y^{2}-2xy-4x+4y-4=0 $
B) $ x^{2}+y^{2}-2xy+4x-4y-4=0 $
C) $ x^{2}+y^{2}+2xy-4x+4y-4=0 $
D) $ x^{2}+y^{2}+2xy-4x-4y+4=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Let focus is $ S(0,0) $ and A is the vertex of parabola. Take any point Z such that AS = AZ. Given tangent at vertex is $ x-y+1=0. $
Since directrix is parallel to the tangent at the vertex.
$ \therefore $ Equation of directrix is $ x-y+\lambda =0 $ , where $ \lambda $ is constant. $ \because $ A is midpoint of SZ,
$ \therefore $ SZ = 2.SA
therefore $ \frac{|0-0+\lambda |}{\sqrt{1^{2}+{{(-1)}^{2}}}}=2\times \frac{|0-0+1|}{\sqrt{1^{2}+{{(-1)}^{2}}}} $
therefore $ |\lambda |=2 $ i.e. $ \lambda =2 $
$ \because $ Directrix in this case always lies in IInd quadrant $ \lambda =2 $
Hence equation of directrix is $ x-y+2=0 $
Now, P be any point on parabola SP = PM
therefore $ SP^{2}=PM^{2} $
therefore $ {{(x-0)}^{2}}+{{(y-0)}^{2}}={{( \frac{|x-y+2|}{\sqrt{2}} )}^{2}} $
therefore $ x^{2}+y^{2}+2xy-4x+4y-4=0. $
BETA
