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