Question

The foci of the conic section $25x^2+16y^2-150x=175$ are

• A. $(0,\pm 3)$
• B. $(0,\pm 2)$
• C. $(3,\pm 3)$
• D. $(0,\pm 1)$

Answer 1

The correct option is C

$(3,\pm 3)$

Explanation for correct option:

Step 1: Find standard equation

Given: $25x^2+16y^2-150x=175$

Using completing the square method we get,

$25(x^2–6x+9–9)+16y^2=175$

Now, ${(x-3)}^2=x^2-6x+9$

So, $25{(x-3)}^2+16y^2=175+225=400$

Therefore,

$\frac{{(\mathrm{x}-3)}^2}{16}+\hspace{0.17em}\frac{{\mathrm{y}}^2}{25}=1$

$\frac{{(\mathrm{x}-3)}^2}{4^2}+\hspace{0.17em}\frac{{\mathrm{y}}^2}{5^2}=1..........\left(1\right)$

From this equation we get to know that the vertex is $(3,0)$.

Step 2: Find the foci of the given conic section

Now ,Comparing the equation (1) with the standard equation of ellipse $\frac{{\mathrm{x}}^2}{b^2}+\hspace{0.17em}\frac{{\mathrm{y}}^2}{a^2}=1$ .

Here , Major axis is $y$ - axis . So , we have $a=5,b=4$

Now, Eccentricity of the ellipse :-

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{e}\mathbf{=}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}} \\ \Rightarrow e=\sqrt{1-\frac{4^2}{5^2}} \\ \Rightarrow e=\sqrt{\frac{25-16}{25}} \\ \Rightarrow e=\sqrt{\frac{9}{25}} \\ \Rightarrow e=\frac{3}{5} \end{gathered}$$

We know that , Coordinates of foci of standard ellipse with major axis as $y$ are $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{ae}\mathbf{)}$

Here , $a=5,e=\frac{3}{5}$

Ellipse is shifted ,so $x-3=0\Rightarrow x=3$

So , foci is $(3,\pm 5\times \frac{3}{5}).$

Foci is $(3,\pm 3)$.

Hence, the foci of the conic section $25x^2+16y^2-150x=175$ are $(3,\pm 3)$.

Therefore, option (C) is the correct answer.