Question

The abscissa of the foci of the ellipse \[25\left( {{x}^{2}}-6x+9 \right)+16{{y}^{2}}=400\] .
A.\[(4,ae)\] , \[(4,-ae)\]
B.\[(3,ae)\] , \[(3,-ae)\]
C.\[(5,ae)\] , \[(5,-ae)\]
D.None of these.

Answer 1

Hint: Transform the equation \[25\left( {{x}^{2}}-6x+9 \right)+16{{y}^{2}}=400\] into the standard equation of an ellipse which is \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] . For the ellipse \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] , the coordinates of the foci are \[x=0\] and \[y=\pm ae\] . Replace x by \[\left( x-3 \right)\] , b by 4, and a by 5 in \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] and the coordinates of the foci are \[x=0\] and \[y=\pm ae\] . We have to take only the x-coordinate of the point because the question is asking for the abscissa and abscissa is the x-coordinate of a point.

Complete step-by-step answer:
According to the question, it is given that the equation of the ellipse is,
\[25\left( {{x}^{2}}-6x+9 \right)+16{{y}^{2}}=400\] …………………………….(1)
Dividing by 400 in LHS and RHS of equation (1), we get
\[\Rightarrow \dfrac{25\left( {{x}^{2}}-6x+9 \right)}{400}+\dfrac{16{{y}^{2}}}{400}=\dfrac{400}{400}\]
\[\Rightarrow \dfrac{\left( {{x}^{2}}-6x+9 \right)}{16}+\dfrac{{{y}^{2}}}{25}=1\] …………………………(2)
We know the formula, \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] ………………………..(3)
Replacing a by x and b by 3 in equation (3), we get
\[{{(x-3)}^{2}}={{x}^{2}}+{{3}^{2}}-2.x.3={{x}^{2}}+9-6x\] ……………………….(4)
Transforming equation (2), we get
\[\Rightarrow \dfrac{{{\left( x-3 \right)}^{2}}}{16}+\dfrac{{{y}^{2}}}{25}=1\] ………………………(5)
The length of x-intercept = \[\sqrt{16}=4\] .
The length of y-intercept = \[\sqrt{25}=5\] .
Here, in this equation, we have the length of the y-intercept more than the length of the x-intercept.
We know the standard equation of the ellipse, \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] …………………….(6)
Here, a is greater than b.
We know that an ellipse has two foci. So, we should have the coordinates of both foci.
For the ellipse \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}}=1\] , the coordinates of the foci are \[x=0\] and \[y=\pm ae\] …………….(7)
Replacing x by \[\left( x-3 \right)\] , b by 4, and a by 5 in equation (7), we get, \[\dfrac{{{\left( x-3 \right)}^{2}}}{16}+\dfrac{{{y}^{2}}}{25}=1\] .
Since an ellipse has two foci so, we should have the coordinates of both foci.
For the ellipse \[\dfrac{{{\left( x-3 \right)}^{2}}}{16}+\dfrac{{{y}^{2}}}{25}=1\] , the coordinates of the foci are \[\left( x-3 \right)=0\] and \[y=\pm ae\] .
Therefore, the coordinates of the foci are \[(3,ae)\] and \[(3,-ae)\] .
The abscissa of a point is its x-axis coordinate.
Here, the x coordinate of both foci is 3.
Hence, the correct option is D.

Note: In this question, one might mark option (B) which is \[(3,ae)\] , \[(3,-ae)\] as an answer which is wrong. \[(3,ae)\] and \[(3,-ae)\] are the coordinates of the foci not the abscissa of foci. The abscissa of a point is its x coordinate.
We can also solve this question with a tricky method. As abscissa is only the x coordinate of a point. Here, every option has two points and every point consists of both x-coordinate and y-coordinate. But, we only need the x-coordinate. So, only option (D) remains which should be the correct one.