Question

Locus of the middle points of all chords of $ \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1 $ , which are at a distance of $ 2 $ units from the vertex of parabola $ {{y}^{2}}=-8ax $ ,

• A. $ {{\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)}^{2}}=\frac{xy}{6} $
• B. $ {{\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)}^{2}}=\left( \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{81} \right) $
• C. $ \left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)=\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4} $
• D. None of the above

Answer 1

Answer: (B)

Let $ (h,\,\,k) $ be the mid-point of a chord of the ellipse $ \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1 $ .
Then, its equation is
$ \frac{hx}{4}+\frac{ky}{9}-1=\frac{{{h}^{2}}}{9}+\frac{{{k}^{2}}}{9}-1 $
$ [\because \,\,T={{S}_{1}}] $
$ \Rightarrow $ $ \frac{hx}{4}+\frac{ky}{9}=\frac{{{h}^{2}}}{4}+\frac{{{k}^{2}}}{9} $
Since, it is a distance of 2 units from the vertex $ (0,\,\,0) $ of the parabola $ {{y}^{2}}=-8ax $ .
$ \therefore $ $ \left| \frac{\frac{{{h}^{2}}}{4}+\frac{{{k}^{2}}}{9}}{\sqrt{\frac{{{h}^{2}}}{16}+\frac{{{k}^{2}}}{81}}} \right|=2 $
$ \Rightarrow $ $ {{\left( \frac{{{h}^{2}}}{4}+\frac{{{k}^{2}}}{9} \right)}^{2}}=4\left( \frac{{{h}^{2}}}{16}+\frac{{{k}^{2}}}{81} \right) $
Hence, the locus of $ (h,\,\,k) $ is $ {{\left( \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9} \right)}^{2}}=4\left( \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{81} \right) $