• # question_answer A chord PQ is a normal to the parabola ${{y}^{2}}=4ax$ at P and subtends a right angle at the vertex. If $SQ=\lambda \,SP$ where S is the focus then the value of $\lambda$ is A)  1                     B)  2 C)  3                                 D)  4

Solution :

We have, $\frac{2}{{{t}_{1}}}.\,\frac{2}{{{t}_{2}}}\,=-1\,\Rightarrow \,{{t}_{1}}{{t}_{2}}\,=-4;$ Also, ${{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}$ ${{t}_{1}}{{t}_{2}}\,=-t_{1}^{2}-2\Rightarrow \,-4+2=-t_{1}^{2}\Rightarrow \,t_{1}^{2}=2$ Also $t_{2}^{2}\,=t_{1}^{2}\,+\frac{4}{t_{1}^{2}}+4\,$ squaring (1) $\Rightarrow \,\,t_{2}^{2}=2+2+4=8$ $SQ=a(1+t_{2}^{2})\,=a(1+8)=9a$ $SP=a(1+t_{1}^{2}\,)=a(1+2)=3a$ $\therefore \,\,\frac{SQ}{SP}=3\Rightarrow \,SQ=3SP\Rightarrow \,\lambda =3$

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