Question 1

The equation of the circle with centre $(0, 2)$ and radius $2$ is:

Accepted Answer

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$. Substituting $h = 0$, $k = 2$, and $r = 2$, we get $(x - 0)^2 + (y - 2)^2 = 2^2$, which simplifies to $x^2 + y^2 - 4y = 0$.

Question 2

Find the coordinates of the focus of the parabola $y^2 = 12x$.

Accepted Answer

The given equation is $y^2 = 12x$. Comparing this with the standard equation $y^2 = 4ax$, we get $4a = 12$, which gives $a = 3$. The focus of the parabola $y^2 = 4ax$ is given by $(a, 0)$. Therefore, the focus is $(3, 0)$.

Question 3

The length of the latus rectum of the parabola $x^2 = -16y$ is:

Accepted Answer

The length of the latus rectum of a parabola of the form $x^2 = -4ay$ is given by $|4a|$. Comparing $x^2 = -16y$ with $x^2 = -4ay$, we find $4a = 16$. Thus, the length of the latus rectum is $16$ units.

Question 4

The eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is:

Accepted Answer

For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a^2 = 25$ and $b^2 = 9$. Thus, $e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Question 5

The foci of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ are:

Accepted Answer

For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $a^2 = 16$ and $b^2 = 9$. The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$. The foci are given by $(\pm ae, 0)$. Here $ae = 4 	imes \frac{5}{4} = 5$. Thus, the foci are $(\pm 5, 0)$.

Conic Sections — CBSE Class 11 Mathematics Important Questions & Revision Notes