Question

HELLLPPP!!!

Answer 1

Given:

Focus of a parabola = $\left(1,\dfrac{1}{2}\right)$

Directrix: $y=3$

To find:

The equation of the parabola.

Solution:

The equation of a vertical parabola is:

$y-k=\dfrac{1}{4a}(x-h)^2$            ...(A)

Where, $(h,k)$ is center, $(h,k+a)$ is focus and $y=k-a$ is the directrix.

On comparing the focus, we get

$(h,k+a)=\left(1,\dfrac{1}{2}\right)$

$h=1$

$k+a=\dfrac{1}{2}$           ...(i)

On comparing the directrix, we get

$k-a=3$              ...(ii)

Adding (i) and (ii), we get

$2k=\dfrac{7}{2}$

$k=\dfrac{7}{4}$

Putting $k=\dfrac{7}{4}$ is (i), we get

$\dfrac{7}{4}+a=\dfrac{1}{2}$

$a=\dfrac{1}{2}-\dfrac{7}{4}$

$a=\dfrac{-5}{4}$

Putting $a=\dfrac{-5}{4},h=1,k=\dfrac{7}{4}$ in (A), we get

$y-\dfrac{7}{4}=\dfrac{1}{4\times \dfrac{-5}{4}}(x-1)^2$

$y-\dfrac{7}{4}=\dfrac{-1}{5}(x^2-2x+1)$

$y-\dfrac{7}{4}=-\dfrac{1}{5}(x^2)-\dfrac{1}{5}(-2x)-\dfrac{1}{5}(1)$

$y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x-\dfrac{1}{5}+\dfrac{7}{4}$

On further simplification, we get

$y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x+\dfrac{35-4}{20}$

$y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x+\dfrac{31}{20}$

Therefore, the equation of the parabola is $y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x+\dfrac{31}{20}$.

Note: Option C is correct but the leading coefficient should be negative.