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

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More