NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

Focus of the parabola $x^2 + y^2 - 2xy - 4(x + y - 1) = 0$ is:

Choose the correct answer:

Explanation

Solution

1. Rearranging the Equation:

x^2 + y^2 - 2xy - 4x - 4y + 4 = 0

(x - y)^2 = 4(x + y - 1)

2. Identifying Axis and Tangent at Vertex:

Axis: $x - y = 0 \implies y = x$
Tangent at Vertex: $x + y - 1 = 0$

3. Finding Vertex ( $V$ ):

Intersection of $y = x$ and $x + y - 1 = 0$ :

x + x - 1 = 0 \implies 2x = 1 \implies x = 1/2, y = 1/2

V = (1/2, 1/2)

4. Finding Focal Length ( $a$ ):

Length of Latus Rectum ( $4a$ ) for $(x-y)^2 = 4(x+y-1)$ :

4a = \frac{4}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}

a = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}

5. Calculating Focus ( $S$ ):

Focus lies on the axis $y=x$ at distance $a$ from Vertex:

x_s = x_v + a \cos 45^\circ = \frac{1}{2} + \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) = \frac{1}{2} + \frac{1}{2} = 1

y_s = y_v + a \sin 45^\circ = \frac{1}{2} + \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\right) = \frac{1}{2} + \frac{1}{2} = 1

Conclusion:

Focus is $(1, 1)$ .

Correct Option:

(a) (1,1)

Choose the correct answer:

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