A) \[{{a}^{2}}=1,{{b}^{2}}=25\]
B) \[{{a}^{2}}=25,{{b}^{2}}=20\]
C) \[{{a}^{2}}=20,{{b}^{2}}=25\]
D) None of these
Correct Answer: D
Solution :
[d] Centre is (1, 2) and focus is (6, 2), hence the line joining the centre C and the focus S (i.e., the axis) is parallel to x-axis. Therefore, the equation must be of the form |
\[\frac{{{(x-1)}^{2}}}{{{a}^{2}}}+\frac{{{(y-2)}^{2}}}{{{b}^{2}}}=1\] ?. (i) |
The distance between the centre and the focus \[CS=ae=6-1=5\] |
Also |
\[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})={{a}^{2}}-{{a}^{2}}{{e}^{2}}={{a}^{2}}-25\] |
\[\therefore \] The equation (i) becomes |
\[\frac{{{(x-1)}^{2}}}{{{a}^{2}}}+\frac{{{(y-2)}^{2}}}{{{a}^{2}}-25}=1\] ?. (ii) |
The point (4, 6) lies on it, therefore |
\[\frac{{{(4-1)}^{2}}}{{{a}^{2}}}+\frac{{{(6-2)}^{2}}}{{{a}^{2}}-25}=1\Rightarrow {{a}^{2}}=45\] |
\[\therefore \,\,\,\,{{b}^{2}}=45-25=20.\] |
The required equation is |
\[\frac{{{(x-1)}^{2}}}{45}+\frac{{{(y-2)}^{2}}}{20}=1.\] |
\[9ac-9{{a}^{2}}-2{{c}^{2}}>9ac-6{{c}^{2}}\] ? (i) |
Again \[3a<2c\Rightarrow 9ac<6{{c}^{2}}\] |
\[\Rightarrow 9ac-6{{c}^{2}}<0\] ? (ii) |
From (i) and (ii), \[9ac-9{{a}^{2}}-2{{c}^{2}}>0\] |
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