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Maths-

General

Easy

Question

Let A be the vertex and L the length of the latus rectum of the parabola y² – 2y – 4x – 7 = 0. The equation of the parabola with A as vertex, 2L the length of the latus rectum and the axis at right angles to that of the given curve is -

x² + 4x + 8y – 4 = 0
x² + 4x + 8y – 12 = 0
x² + 4x + 8y + 12 = 0
x² + 8x – 4y + 8 = 0

hint Hint:

Let (y-k)² = 4a(x-h) the (h,k) is the vertex and 4a is the length of latus rectum of the parabola and y = k is the axis of the parabola.

The correct answer is: x² + 4x + 8y – 4 = 0

Given, the equation y² – 2y – 4x – 7 = 0
y²- 2y +1 = 4x + 7+ 1
(y-1)² = 4(x+2)
here length of latus rectum L = 4 , vertex of parabola = (-2,1) and axis of parabola is y = 1
Then equation for perpendicular y = 1 and veterx (-2,1) is x = -2
parabola length of latus rectum = 2L = 2*4 = 8
the equation of parabola with axis x = -2 and latus rectum = 8 and vertex (-2,1) is
$left parenthesis x plus 2 right parenthesis squared space equals plus-or-minus 8 left parenthesis y minus 1 right parenthesis x squared plus 4 x plus 4 space equals 8 y minus 8 space space space space space space space left parenthesis o r right parenthesis space x squared plus 4 x plus 4 space equals negative 8 y plus 8 x squared plus 4 x minus space 8 y plus 8 plus 4 space equals 0 space left parenthesis o r right parenthesis x squared plus 4 x plus space 8 y minus 8 plus 4 space equals 0 x squared plus 4 x minus space 8 y plus 12 space equals 0 space space space space space space left parenthesis o r right parenthesis space x squared plus 4 x plus space 8 y minus 4 equals 0$
Therefore, x² + 4x + 8y – 4 = 0 is the parabola that with stand the coditions.

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