NEB Class 12 Science Mathematics (NEB Released Items) Question Paper 2081 (Set C) Nepal

(a) Ellipse $9x^2 + 4y^2 = 36$.

Divide by $36$: $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.

Here $a^2 = 9$, $b^2 = 4$ with $a^2 > b^2$, so the major axis is along the $y$-axis: $a = 3$, $b = 2$.

• Eccentricity: $e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{4}{9}} = \dfrac{\sqrt{5}}{3}$.
• Foci: $(0, \pm ae) = (0, \pm 3 \cdot \tfrac{\sqrt5}{3}) = (0, \pm\sqrt{5})$.
• Vertices: $(0, \pm a) = (0, \pm 3)$.

(b) Parabola $y^2 - 6y + 9 - 8x = 0$.

Rewrite: $(y - 3)^2 = 8x$.

This is of the form $(y - k)^2 = 4p(x - h)$ with $h = 0$, $k = 3$ and $4p = 8 \Rightarrow p = 2$.

• Vertex: $(h, k) = (0, 3)$.
• Focus: $(h + p, k) = (2, 3)$.

(a) Show $b = 2$.

By the cosine rule, $b^2 = a^2 + c^2 - 2ac\cos B$.

With $a = \sqrt2$, $c = \sqrt3 - 1$, $B = 135^\circ$ so $\cos B = -\dfrac{1}{\sqrt2}$:

$b^2 = 2 + (\sqrt3 - 1)^2 - 2\sqrt2(\sqrt3 - 1)\left(-\dfrac{1}{\sqrt2}\right)$

$(\sqrt3 - 1)^2 = 4 - 2\sqrt3$, and $-2\sqrt2(\sqrt3-1)\cdot(-\tfrac{1}{\sqrt2}) = 2(\sqrt3 - 1) = 2\sqrt3 - 2$.

$b^2 = 2 + (4 - 2\sqrt3) + (2\sqrt3 - 2) = 4$, hence $b = 2$.

(b) Prove $a = b\cos C + c\cos B$ by the vector method.

Let the position vectors of the vertices be $\vec{A}, \vec{B}, \vec{C}$. The sides as vectors satisfy $\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}$, i.e. $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ where $\vec{a} = \vec{BC}$, $\vec{b} = \vec{CA}$, $\vec{c} = \vec{AB}$, with $|\vec a| = a$, $|\vec b| = b$, $|\vec c| = c$.

From $\vec a = -(\vec b + \vec c)$, take the dot product with $\vec a$:

$\vec a \cdot \vec a = -\vec a \cdot \vec b - \vec a \cdot \vec c$.

The interior angle relations give $\vec a \cdot \vec b = -ab\cos C$ and $\vec a \cdot \vec c = -ac\cos B$ (the angle between the directed sides is the supplement of the interior angle).

Therefore $a^2 = ab\cos C + ac\cos B$, and dividing by $a$:

(c) Hyperbola $\dfrac{(x-5)^2}{16} - \dfrac{(y+2)^2}{9} = 1$.

Centre $(5, -2)$, with $a^2 = 16 \Rightarrow a = 4$ and $b^2 = 9 \Rightarrow b = 3$. Transverse axis is horizontal.

• Eccentricity: $e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{9}{16}} = \dfrac{5}{4} = 1.25$.
• Vertices: $(5 \pm a, -2) = (9, -2)$ and $(1, -2)$.
• Foci: $(5 \pm ae, -2)$ with $ae = 4 \cdot \tfrac54 = 5$, giving $(10, -2)$ and $(0, -2)$.