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.