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  • Which of the following is true for the parabola ?<div class=

Which of the following is true for the parabola y^2= 3x +8y + 2?

Option: 1

Vertex is (6,4)


Option: 2

Focus is \left (\frac{27}{4}, 4 \right )


Option: 3

Directrix is x-\frac{27}{4}=0


Option: 4

None of the above is true


Answers (1)

best_answer

 

 

Equation of a Parabola When The Vertex IS (h k ) and Parabolic Curve -

Equation of a Parabola When The Vertex IS (h k ) and Parabolic Curve 

The equation of the parabola when the axis is parallel to x-axis 

\\\mathrm{y^2=4ax}

can be written as \\\mathrm{(y-0)^2=4a(x-0)}

The vertex of the parabola is O(0, 0). Now the origin is shifted to O’(h, k) without changing the direction of axes, its equation becomes \\\mathrm{(y-k)^2=4a(x-h)}


\begin{array}{l}{\text { The parametric equation of the curve }(y-k)^{2}=4 a(x-h)} \\ {\text { are } x-h+a t^{2} \text { and } y=k+2 a t}\end{array}

Thus its focus is S (a + h, k), latus rectum = 4a and the equation of the directrix is

x = h  – a, i.e. x + a – h = 0

-

 

 

The given equation is

\\ \begin{aligned} & y^{2}=3 x+8 y+2 \\ \Rightarrow & y^{2}-4 y=x^{2}+2 \\ \Rightarrow & y^{2}-8 y+16=3x+18 \\ \Rightarrow &(y-4)^{2}=3(x+6) \\ \Rightarrow & Y^{2}=3 X \\ & \end{aligned}

\\\text {where } X=& x+6 \text { and } Y=y-4 \\ \text { Vertex: }(-6,4)

\begin{array}{ll}{\text { Focus: }} {(a, 0)} \\\\ {\Rightarrow } {X=a, Y=0} \\\\ {\Rightarrow } {x+6=\frac{3}{4}, y-4=0} \\\\ {\Rightarrow } {x=-\frac{21}{4} \text { and } y=4} \\\\ {\text { Hence, the focus is }\left(-\frac{21}{4}, 4\right)}\end{array}

\begin{array}{ll}{\text { Latus rectum: } 4 a=3} \\ {\text { Directrix: } X+a=0} \\ {\Rightarrow \quad x+6=3 / 4} \\ {\Rightarrow \quad x=-21 / 3} \\ {\Rightarrow \quad} {3 x+21=0} \\ {\text { Axis: }} {Y=0} \\ {\Rightarrow\quad} {y-4=0} \\ {\Rightarrow\quad} {y=4}\end{array}

Hence none of the above is true

Posted by

Ajit Kumar Dubey

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