Conics

Geometry Level 4

Find the coordinates of the vertex of the conic represented as x 2 + y 2 − 2 x y + 4 x − 2 y + 1 = 0 x^2 +y^2 -2xy +4x -2y +1 =0 .

( − 1 4 , 5 8 ) \left( -\frac14, \frac{5}8 \right) ( − 1 8 , 11 8 ) \left( -\frac18, \frac{11}8 \right) ( 1 8 , − 11 4 ) \left( \frac18, -\frac{11}4 \right) ( 1 8 , − 13 6 ) \left( \frac18, -\frac{13}6 \right) ( 1 4 , − 13 8 ) \left( \frac14, -\frac{13}8 \right)

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Tanishq Varshney
Jan 26, 2015

Rewriting the equation ( x − y ) 2 = − 4 x + 2 y − 1 (x-y)^2 =-4x +2y -1 The given equation is of a parabola as h 2 = a b h^2 =ab According to definition of parabola. (Length of perpendicular on axis of parabola)^2 = (Length of latus rectum)(Length of perpendicular on tangent at the vertex)

Also Tangent at the vertex and axis of parabola are perpendicular. But in the above equation it is not so, we introduce a constant k k . ( x − y + k ) 2 = − 4 x + 2 y − 1 + 2 k x − 2 k y + k 2 (x-y+k)^2 =-4x +2y -1 +2kx -2ky +k^2 [actually we have added and subtracted the required terms] ( x − y + k ) 2 = 2 ( k − 2 ) x + 2 ( 1 − k ) y + k 2 − 1 (x-y+k)^2 =2(k-2)x +2(1-k)y +k^2 -1 .

For perpendicular lines m 1 m 2 = − 1 m_1 m_2=-1

1 − 2 ( k − 2 ) 2 ( 1 − k ) = − 1 1\frac{-2(k-2)}{2(1-k)} =-1

On solving, k = 3 2 k= \frac{3}{2}

( x − y + 3 2 ) 2 = − ( x + y − 5 4 ) (x-y+\frac{3}{2})^2 = -( x+y- \frac{5} {4}) Further finding the point of intersection of the lines gives the vertex.

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Why can't we just partially differentiate the conic and get the vertex

Swapnil Vatsal - 3 years, 6 months ago

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