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Question

Find the equation of the parabola which is symmetric about the $$\displaystyle y$$-axis and passes through the point $$\displaystyle P(2, -3)$$.


A
3y2=4x
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B
3y2=4x
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C
y2=4x
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D
None of these
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Solution

The correct option is B $$3{ y }^{ 2 }=-4x$$
When the parabola is symmetrical about y-axis, its general equation is given by,
$${ x }^{ 2 }=4ay$$ or $${ x }^{ 2 }=-4ay$$

Now, point $$P\left( 2,-3 \right) $$ lies in forth quadrant and hence, parabola will open up downwards.

Thus, equation of parabola will be $${ x }^{ 2 }=-4ay$$

Now, point P must satisfy equation of parabola as it lies on parabola.
$$\therefore { \left( 2 \right)  }^{ 2 }=-4a\left( -3 \right) $$
$$\therefore 4=12a$$
$$\therefore a=\frac { 4 }{ 12 } $$
$$\therefore a=\frac { 1 }{ 3 } $$

Put this value in equation of parabola, we get,
$${ y }^{ 2 }=-4\left( \frac { 1 }{ 3 }  \right) x$$
$$\therefore { y }^{ 2 }=\left( \frac { -4 }{ 3 }  \right) x$$
$$\therefore 3{ y }^{ 2 }=-4x$$

Maths

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