Question bank/JEE Main/Mathematics

A variable chord PQ of y^2= 4ax subtends a right angle at its
vertex then the locus of point of
intersection of normals at P and Q will be
ijoy; y^2= 4ax dh,d pj thok PQ tks bl ij ledks.k cukrh

Ay^2= 4a(x - a)By^2= 8a(x - 4a)Cy^2= 12a(x - 8a)Dy^2= 16a(x - 6a)

Show answer and solution▾

$\frac{1}{t_{1}} \cdot \frac{2}{t_{2}} = - 1$
$t_{1}t_{2} = - 4$
Normals at P and Q are
$y + t_{1}x = 2at_{1} + at_{1}^{3}$
$y + t_{2}x = 2at_{2} + at_{2}^{3}$
point of intersection of normals will be
$x = 2a + a\left( t_{1}^{2} + t_{2}^{2} + t_{1}t_{2} \right) = a\left( t_{1} + t_{2} \right)^{2} + 6a$
$y = - at_{1}t_{2}\left( t_{1} + t_{2} \right) = 4a\left( t_{1} + t_{2} \right)$
$(x - 6a) = a\left( \frac{y^{2}}{16a^{2}} \right)$
$y^{2} = 16a(x - 6a)$

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx