Question bank/JEE Main/Mathematics

The base of a triangle passes through a fixed point (f, g) and
its sides are bisected at right angles by the lines y^2 - 8xy - 9x^2
= 0. The locus of vertex of triangle is,d f=Hkqt,d fLFkj fcUnq (f, g) ls xqtjrk gS rFkk Hkqtk, y^2 - 8xy - 9x^2 = 0 ls ledksa.k ij lef}Hkkftr gksrh

AStraight line ljy js[kkBCircle o`ÙkCellipse nh?kZo`Ùk

Show answer and solution▾

equation of BC is BC dk lehdj.k
y + h = $\frac{50h + 40k}{50k - 40h}$ (x + k)
(f, g) lies on BC ij $\Rightarrow$ g + h = $\frac{5h + 4k}{5k - 4h}$
(f + k)
$\Rightarrow$ locus of (h, k) dk fcUnqiFk is 4x^2 + 4y^2 + x (5f +
4g) + y (4f - 5g) = 0 gSA

Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx