• # question_answer The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse ${{x}^{2}}+2{{y}^{2}}=2$ between the co-ordinates axes, is                                                      [IIT Screening 2004] A)            $\frac{1}{{{x}^{2}}}+\frac{1}{2{{y}^{2}}}=1$                               B)            $\frac{1}{4{{x}^{2}}}+\frac{1}{2{{y}^{2}}}=1$ C)            $\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1$                             D)            $\frac{1}{2{{x}^{2}}}+\frac{1}{{{y}^{2}}}=1$

Solution :

Let the point of contact be            $R\equiv (\sqrt{2}\cos \theta ,\,\sin \theta )$            Equation of tangent AB is            $\frac{x}{\sqrt{2}}\cos \theta +y\sin \theta =1$            $\Rightarrow$$A\equiv (\sqrt{2}\sec \theta ,\,0);\,B\equiv (0,\,\text{cosec }\theta )$            Let the middle point Q of AB be $(h,\,k)$            $\Rightarrow$ $h=\frac{\sec \theta }{\sqrt{2}},\,k=\frac{\text{cosec }\theta }{2}\Rightarrow \cos \theta =\frac{1}{h\sqrt{2}},\,\sin \theta =\frac{1}{2k}$            $\Rightarrow$ $\frac{1}{2{{h}^{2}}}+\frac{1}{4{{k}^{2}}}=1$, \Required locus is $\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1$.            Trick :  The locus of mid-points of the portion of tangents to the ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ intercepted between axes is ${{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}=4{{x}^{2}}{{y}^{2}}$            i.e., $\frac{{{a}^{2}}}{4{{x}^{2}}}+\frac{{{b}^{2}}}{4{{y}^{2}}}=1$ or $\frac{1}{2{{x}^{2}}}+\frac{1}{4{{y}^{2}}}=1$.

You need to login to perform this action.
You will be redirected in 3 sec