A) \[\frac{{{x}^{2}}}{5}+\frac{{{y}^{2}}}{8}=1\]
B) \[\frac{{{x}^{2}}}{40}+\frac{{{y}^{2}}}{10}=10\]
C) \[\frac{{{x}^{2}}}{40}+\frac{{{y}^{2}}}{10}=1\]
D) \[\frac{{{x}^{2}}}{10}+\frac{{{y}^{2}}}{40}=1\]
Correct Answer: C
Solution :
Let the equation of ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] ? (i) We know that, the line\[y=mx+c\]is a tangent to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], if\[{{c}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}\] It is given that the line\[3x-2y-20=0\] \[ie,\]\[y=\frac{3}{2}x-10\]is a tangent to the ellipse \[(i)\] \[\therefore \] \[{{(-10)}^{2}}={{a}^{2}}{{\left( \frac{3}{2} \right)}^{2}}+{{b}^{2}}\] \[\Rightarrow \] \[\frac{9}{4}{{a}^{2}}+{{b}^{2}}=100\] \[\Rightarrow \] \[9{{a}^{2}}+4{{b}^{2}}=400\] ... (ii) Similarly,\[x+6y-20=0\] \[ie,\]\[y=-\frac{1}{6}x+\frac{10}{3}\]is also a tangent to the ellipse (i) \[\therefore \] \[{{\left( \frac{10}{3} \right)}^{2}}={{a}^{2}}{{\left( -\frac{1}{6} \right)}^{2}}+{{b}^{2}}\] \[\Rightarrow \] \[\frac{1}{36}{{a}^{2}}+{{b}^{2}}=\frac{100}{9}\] \[\Rightarrow \] \[{{a}^{2}}+36{{b}^{2}}=400\] ... (iii) On solving Eq. (ii) and (iii), we get \[{{a}^{2}}=40,\,\,{{b}^{2}}=10\] Putting these values in Eq. (i), we get the required equation of ellipse \[\frac{{{x}^{2}}}{40}+\frac{{{y}^{2}}}{10}=1\]
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