Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the area of $\Delta ABC$ where AC is focal chord of the parabola $y^2=4ax$ & B is the vertex of parabola $y^2=4ax$?

$\begin{array}{1 1}(a)\;a^2|t_1t_2(t_1-t_2)|\\(b)\;a|t_1t_2(t_1t_2)|\\(c)\;2a^2|t_1t_2(t_1t_2)|\\(d)\;a^2|t_1t_2(t_1+t_2)|\end{array}$

Can you answer this question?

1 Answer

0 votes
Vertex B(0,0)
For $\Delta ABC$,$A(at_1^2,2at_1),B(0,0),C(at_2^2,2at_2)$.
Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
Where $s=\large\frac{a+b+c}{2}$
$a$- distance between AB.
$b$- distance between BC
$c$- distance between AC.
Area of $\Delta ABC=\large\frac{1}{2}$$\begin{vmatrix}at_1^2&2at_1&1\\at_2^2&2at_2&1\\0&0&1\end{vmatrix}$
$\Rightarrow \large\frac{1}{2}$$\mid 2a^2t_1^2t_2-2a^2t_1t_2^2\mid$
$\Rightarrow \large\frac{1}{2}$$\times 2\mid a^2t_1^2t_2-a^2t_1t_2^2\mid$
$\Rightarrow a^2\mid t_1t_2-t_1t_2^2\mid$
$\Rightarrow a^2\mid t_1t_2(t_1-t_2)\mid$
Hence (a) is the correct answer.
answered Feb 5, 2014 by sreemathi.v
edited Feb 5, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App