Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i) $\dfrac{x^2}{25}+\dfrac{y^2}9=1$

(ii) $\dfrac{x^2}3+\dfrac{y^2}{10}=1$

(iii) $\dfrac{x^2}{25}-\dfrac{y^2}{144}=1$

(iv) $\dfrac{y^2}{16}-\dfrac{x^2}9=1$

Question

Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i) $\dfrac{x^2}{25}+\dfrac{y^2}9=1$

(ii) $\dfrac{x^2}3+\dfrac{y^2}{10}=1$

(iii) $\dfrac{x^2}{25}-\dfrac{y^2}{144}=1$

(iv) $\dfrac{y^2}{16}-\dfrac{x^2}9=1$

Accepted Answer

> [!TLDR]
>
> Each equation is already in standard form; identify ellipse vs. hyperbola from the sign between the terms, then read $a^2,b^2$ and compute $c$ (via $c^2=a^2-b^2$ for the ellipse, $c^2=a^2+b^2$ for the hyperbola).

**Part (i). $\frac{x^2}{25}+\frac{y^2}9=1$ — ellipse** ($25>9$, major axis $x$). $a^2=25,b^2=9\Rightarrow a=5,b=3$. $c^2=25-9=16\Rightarrow c=4$. Centre $(0,0)$; foci $(\pm4,0)$; vertices $(\pm5,0)$; directrices $x=\pm\frac{a^2}c=\pm\frac{25}4$.

**Part (ii). $\frac{x^2}3+\frac{y^2}{10}=1$ — ellipse** ($10>3$, major axis $y$). $a^2=10,b^2=3$. $c^2=10-3=7\Rightarrow c=\sqrt7$. Centre $(0,0)$; foci $(0,\pm\sqrt7)$; vertices $(0,\pm\sqrt{10})$; directrices $y=\pm\frac{10}{\sqrt7}$.

**Part (iii). $\frac{x^2}{25}-\frac{y^2}{144}=1$ — hyperbola** (positive $x^2$ term, transverse axis $x$). $a^2=25,b^2=144\Rightarrow a=5,b=12$. $c^2=25+144=169\Rightarrow c=13$. Centre $(0,0)$; foci $(\pm13,0)$; vertices $(\pm5,0)$; directrices $x=\pm\frac{25}{13}$. …

Condition	Type
$A=C\ne0$	Circle
exactly one of $A,C$ is $0$	Parabola
$A\ne C$ , same sign	Ellipse

Q.Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

Concept understanding — Conic as a Section / General Second-Degree Equation

Q.Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

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