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Parabola

Section: Mathematics

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Question : 4 of 15

Marks: +1, -0

If the perpendicular distance from the focus of a parabola

y^2=4ax

to its directrix is

\;\frac{3}{2}

, then the equation of the normal drawn at

(4 a,-4 a)

[AP EAPCET 22 May 2025 S2]

$2x+y=3$
$2x-y=9$
$x-2y=9$
$x+2y+3=0$

Solution:

y^2=4ax,\;\text{ focus }\;=(a,0)

Equation of directrix is

x=-a

.

The perpendicular distance from the focus

(a,0)

to the directrix

x+a=0

\;\Rightarrow\;\; \frac{\left|a+a\right|}{\sqrt{1+0}}=\left|2a\right|=\left|2a\right|

\;\Rightarrow\;\; \left|2a\right|=\;\frac{3}{2}

\;\Rightarrow\;\; 2a=\;\frac{3}{2}\;\text{ or }\;2a=\;\frac{-3}{2}

\;\Rightarrow\;\; 2a=\;\frac{3}{2}\Rightarrow a=\;\frac{3}{4}

\;\Rightarrow\;\; y-y_1=-\;\frac{y_1}{2a}(x-x_1)

Point

(4 a,-4 a)

\;x_1=4a, y_1=-4a

\Rightarrow y-(-4a)=-\frac{(-4a)}{2a}(x-4a)

\Rightarrow y+4a=-(-2)(x-4a)

\Rightarrow y+4a=2(x-4a)

\Rightarrow y+4a=2x-8a

\Rightarrow 2x-y-12a=0

\;\;\text{ Substituting }\; a=\;\frac{3}{4}

\;\Rightarrow\;\; 2x-y-12\left(\;\frac{3}{4}\right)=0

\;\Rightarrow\;\; 2x-y-9=0

\;\;2x-y=9

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