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Q2.

Expert-verifiedFound in: Page 423

Book edition
Middle English Edition

Author(s)
Carter

Pages
804 pages

ISBN
9780079039903

**Write an equation of the parabola that will be open to the left.**

The vertex is $\left(3,-7\right)$. Focus is $\left(3,-\frac{111}{16}\right)$. Axis of symmetry is $x=3$. The equation of directrix is $y=-\frac{113}{16}$.

The given equation is $y=4{\left(x-3\right)}^{2}-7$.

For two different forms of equations of parabola stated below, use the following key-concept to find vertex, axis of symmetry, focus, directrix, direction of opening of parabola and length of latus rectum.

From the concept stated above, the parabola will be of the form $x=a{\left(y-k\right)}^{2}+h$ where $a<0$. Therefore, the equation of the parabola is written as:

$\begin{array}{c}x=a{\left(y-k\right)}^{2}+h\mathrm{....}\left(\begin{array}{l}\text{where}a<0\text{and}\\ \left(h,k\right)\in R\text{}\left(\text{Real}\right)\end{array}\right)\\ x=-3{\left(y-3\right)}^{2}+\mathrm{4....}\left(\begin{array}{l}\text{where}a=-3\text{and}\\ h=4,k=-3\end{array}\right)\end{array}$

The equation of the parabola which will be open to left is $x=-3{\left(y-3\right)}^{2}+4$.

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