Question

find the equation of parabola whose focus is (1,-1) and whose vertex is 2,1 also find its axis and latus rectum​

Answer 1

Answer:

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Answer 2

$4x^2 + y^2 + 89 - 4xy - 34y - 32x = 0$ is Parabolic Equation

Step-by-step explanation:

First of all point (-1,1) lie in IInd Quadrant which means the equation of parabola will be  

$y^2 = - 4ax$

thus,  

(1,2) = (x,y)

Putting the above values,  

$(2)^2 = - 4a(1)$

$\frac {4}{-4} = a$

Hence, Result is, Value of a is -1.

And, Finally, Latus Rectum is = 4a = $4 \times (-1)$

= -4

Also,  

Line Joining (1,-1) with (2,1) has a Parabola Axis as y=2x-3  

With latus rectum = $4 \times (Distance\ between\ focus\ and\ vertex)$

Or, $4 \times (5^{(\frac {1}{2})})$

= $4 \times 2.236$

= 8.944

So, $4x^2 + y^2 + 89 - 4xy - 34y - 32x = 0$ can be called as equation of parabola  

(From Parabola Definition)