Browse Questions

# Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the following : $y^2=10x$

$\begin {array} {1 1} (A)\;focus \bigg(\large\frac{5}{2}, 0\bigg) , Axis : x - axis, Equation \: of \: directrix : 2x+5=0, Length \: of \: the \: latus \: rectum : 10 \\ (B)\; focus \bigg( 0, \large\frac{5}{2} \bigg) , Axis : y - axis, Equation \: of \: directrix : 2y+5=0, Length \: of \: the \: latus \: rectum : 10 \\ (C)\;focus \bigg(-\large\frac{5}{2}, 0\bigg) , Axis : x - axis, Equation \: of \: directrix : 2x-5=0, Length \: of \: the \: latus \: rectum : 10 \\ (D)\;focus \bigg(0, -\large\frac{5}{2} \bigg) , Axis : y - axis, Equation \: of \: directrix : 2y-5=0, Length \: of \: the \: latus \: rectum : 10 \end {array}$

Toolbox:
• Equation of a parabola is $y^2=4ax$
• If x is positive , then the parabola opens towards the right.
• If the equation involves $y^2$ then the axis of the parabola is x - axis.
• Equation of the directrix is x = -a or x+a=0
• Length of the lactus rectum is 4a.
Step 1 :
The given equation is $y^2 = 10x$
Here the coefficient of x is positive.
Hence the parabola opens towards the right.
Comparing this equation with
$y^2=4ax$
$\Rightarrow 4a=10$
$\Rightarrow a = \large\frac{10}{4}$ or $\large\frac{5}{2}$
$\therefore$ Coordinates of the focus = $\bigg( \large\frac{5}{2}$$, 0 \bigg) Since the given equation involves y^2 , the axis of the parabola is x - axis. Axis : x - axis. Equation of the directrix is x = -\large\frac{5}{2} or x+\large\frac{5}{2}$$=0$
$\Rightarrow 2x+5=0$
Length of the latus rectum is $4 \times \large\frac{5}{2}$$=10$