Question

Find the equation of parabola whose vertex is (0,0) and passing through (5,2) and symmetric with respect to y axis.

Answer 1

Hint: The given vertex is (0, 0) and the given parabola is symmetric with respect to y-axis. Equation of parabola is either of the form ${{x}^{2}}$=4ay or ${{x}^{2}}$=-4ay. Parabola passes through point (5,2) which lies in the first quadrant. Therefore the equation of parabola is of the form ${{x}^{2}}$=4ay .

Complete step-by-step answer:

The parabola passes through point(5,2) it must satisfy the equation ${{x}^{2}}$=4ay-(1)

$25\, = \,(4{\text{a}})\,2$

${\text{a}}\,{\text{ = }}\,\dfrac{{25}}{8}$

Substitute value of a in eqn (1)

${{\text{x}}^{\text{2}}}\,{\text{ = }}\,{\text{4}}\,{{ \times }}\,\dfrac{{{\text{25}}}}{{\text{8}}}\,{{ \times }}\,{\text{y}}$

${{\text{x}}^{\text{2}}}\,{\text{ = }}\,\dfrac{{{\text{25}}}}{{\text{2}}}{\text{y}}$

Hence equation of parabola is $2{x^2}$=25y

Note: We should not be confused whether the equation is

${{\text{x}}^{\text{2}}}\,{\text{ = }}\,{\text{4ay}}$

${\text{(or)}}\,{{\text{x}}^{\text{2}}}\,{\text{ = }}\,{\text{ - 4ay}}$

we can confirm it by drawing the rough figure and putting the given point.