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Parabola

Section: Mathematics

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Question : 2 of 15

Marks: +1, -0

If the tangents of the parabola

y^2=8 x

passing through the point

P(1,3)

touches the parabola at

A

B

, then the area (in sq. units) of

\triangle ABC

[AP EAPCET 23 May 2025 S2]

1
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

y^2=8 x, a=2

Equation of tangent

y=mx+\frac{2}{m}

\text{ at }(1,3)

3=m+\frac{2}{m} \Rightarrow m^2-3m+2=0

\Rightarrow m=1,2

So, equation of tangent (put point

(1,3)

on

y=mc+c

)

y=x+c, y=2x+c

\Rightarrow y=x+2 y=2x+1

to find coordinates of

A

and

B

, put the line on parabola.

(x+2)^2=8x \Rightarrow x^2-4x+4=0

\Rightarrow x=2,-2

\text{ at } x=2, y=\sqrt{8\times2}=4

So,

A=(2,4)

and

(2x+1)^2=8x

\Rightarrow 4x^2-4x+1=0

\Rightarrow x=\frac{1}{2},-\frac{1}{2}

\text{ at } x=\frac{1}{2}, y=2

So,

B=\left(\frac{1}{2},2\right)

Now, area of

\triangle PAB

A=\frac{1}{2}\begin{vmatrix} 1 & 3 & 1 \\ 2 & 4 & 1 \\ \frac{1}{2} & 2 & 1 \end{vmatrix}

=\left| \frac{1}{2} \left(1(4-2)-3\left(2-\frac{1}{2}\right)+1(4-2)\right) \right|..

=\left| \frac{1}{2} \left(2-\frac{9}{2}+2\right) \right|=\left| \frac{1}{2} \left(4-\frac{9}{2}\right) \right|

=\left| \frac{1}{2} \left(-\frac{1}{2}\right) \right|=\frac{1}{4}

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