Question

What are foci of the conic section \[25{x^2} + 16{y^2} - 150x = 175\] ?
A. \[\left( {0, \pm 3} \right)\]
B. \[\left( {0, \pm 2} \right)\]
C. \[\left( {3, \pm 3} \right)\]
D. \[\left( {0, \pm 1} \right)\]

Answer 1

Hint: The given equation represents an ellipse. Rewrite the equation in its standard form \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{a^2}}} = 1\], where \[a > b\]. Then find its foci \[\left( {\alpha ,\beta \pm ae} \right)\] after comparing the given equation with the standard form.

Formula Used:
Standard form of an ellipse is \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{b^2}}} = 1\] or \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{a^2}}} = 1\], where \[a > b\]
Eccentricity of the ellipse is \[e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \]
Foci of the ellipse are \[\left( {\alpha ,\beta \pm ae} \right)\]

Complete step-by-step answer:
The given equation is \[25{x^2} + 16{y^2} - 150x = 175\]
This is a quadratic equation in two variables \[x\] and \[y\], in which coefficient of \[{x^2}\] and coefficient of \[{y^2}\] are not equal but both of them is positive and coefficient of \[xy\] is zero. So, it represents an ellipse.
Standard form of an ellipse is \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{b^2}}} = 1\] or \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{a^2}}} = 1\], where \[a > b\]
Convert the given equation into this standard form.
\[25{x^2} + 16{y^2} - 150x = 175\]
Write the terms in \[x\] first and then in \[y\].
\[ \Rightarrow 25{x^2} - 150x + 16{y^2} = 175\]
Take \[25\] common from the first two terms.
\[ \Rightarrow 25\left( {{x^2} - 6x} \right) + 16{y^2} = 175\]
Make the term within first bracket a perfect square. Add \[9\] to the terms within bracket and \[9 \times 25 = 225\] to the constant term to balance.
\[ \Rightarrow 25\left( {{x^2} - 6x + 9} \right) + 16{y^2} = 400\]
Use the identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow 25{\left( {x - 3} \right)^2} + 16{y^2} = 400\]
Divide all the terms by \[400\] to make the right-hand side \[1\]
\[ \Rightarrow \dfrac{{{{\left( {x - 3} \right)}^2}}}{{16}} + \dfrac{{{{\left( {y - 0} \right)}^2}}}{{25}} = 1\]
Comparing the above equation with the standard form \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{a^2}}} = 1\], we get
\[\alpha = 3,\beta = 0,a = 5,b = 4\]
Eccentricity of the ellipse is \[e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \]
Putting \[a = 5\] and \[b = 4\], we get
\[e = \sqrt {1 - \dfrac{{16}}{{25}}} = \sqrt {\dfrac{9}{{25}}} = \dfrac{3}{5}\]
So, foci of the ellipse are \[\left( {\alpha ,\beta \pm ae} \right)\]
Here \[\alpha = 3,\beta = 0,a = 5,e = \dfrac{3}{5}\]
Now, \[\beta \pm ae = 0 \pm 5 \times \dfrac{3}{5} = \pm 3\]
\[\therefore \]Foci of the ellipse are \[\left( {3, \pm 3} \right)\]
Hence option C is correct.

Note: Whenever the denominator of \[{\left( {x - \alpha } \right)^2}\] will be greater than the denominator of \[{\left( {y - \beta } \right)^2}\], you must choose the standard form \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{b^2}}} = 1\] to compare with the given equation and in this case foci are \[\left( {\alpha \pm ae,\beta } \right)\]. Here the denominator of \[{\left( {x - \alpha } \right)^2}\] is less than the denominator of \[{\left( {y - \beta } \right)^2}\]. So, the given equation has been compared with the standard form \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{b^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{a^2}}} = 1\] and the foci are \[\left( {\alpha ,\beta \pm ae} \right)\].