question_answer

The equation of the ellipse with its centre at \[(1,2),\] focus at (6,2) and passing through the point \[(4,6)\]is \[\frac{{{(x-1)}^{2}}}{{{a}^{2}}}+\frac{{{(y-2)}^{2}}}{{{b}^{2}}}=1\], then

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), \[9ac-9{{a}^{2}}-2{{c}^{2}}>0\]