• # question_answer The locus of mid point of that chord of parabola which subtends right angle on the vertex will be   [UPSEAT 1999] A)            ${{y}^{2}}-2ax+8{{a}^{2}}=0$  B)            ${{y}^{2}}=a(x-4a)$ C)            ${{y}^{2}}=4a(x-4a)$                D)            ${{y}^{2}}+3ax+4{{a}^{2}}=0$

Equation of parabola ${{y}^{2}}=4ax$                       ?..(i)            Equation of that chord of parabola whose mid point is $({{x}_{1}},\,{{y}_{1}})$ will be $y{{y}_{1}}-2a(x+{{x}_{1}})=y_{1}^{2}-4a{{x}_{1}}$            or $y{{y}_{1}}-2ax=y_{1}^{2}-2a{{x}_{1}}$ or $\frac{y{{y}_{1}}-2ax}{y_{1}^{2}-2a{{x}_{1}}}=1$    .....(ii)            Making equation (i) homogeneous by equation (ii), the equation of lines joining the vertex (0,0) of parabola to the point of intersection of chord (ii) and parabola (i) will be            ${{y}^{2}}=4ax\frac{y{{y}_{1}}-2ax}{y_{1}^{2}-2a{{x}_{1}}}$ or ${{y}^{2}}(y_{1}^{2}-2a{{x}_{1}})=4ax(y{{y}_{1}}-2ax)$            or $8{{a}^{2}}{{x}^{2}}-4a{{y}_{1}}xy+(y_{1}^{2}-2a{{x}_{1}}){{y}^{2}}=0$            If lines represented by it are mutually perpendicular, then coefficient of ${{x}^{2}}+$ coefficient of ${{y}^{2}}=0$            therefore, $8{{a}^{2}}+(y_{1}^{2}-2a{{x}_{1}})=0$ or $y_{1}^{2}-2a{{x}_{1}}+8{{a}^{2}}=0$.                    \ Required locus of $({{x}_{1}},\,{{y}_{1}})$ is ${{y}^{2}}-2ax+8{{a}^{2}}=0.$