2 Marks Questions

Question 12 Marks

If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.

Answer

We have circle through the point A(0, 0), B(a, 0) and C(0, b).
Clearly triangle is right angled at vertex A.

So, centre of the circle is the mid point of hypotenuse BC which is $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$

View full question & answer→

Question 22 Marks

Find the equation of the hyperbola with:
$\text{Vertices }(\pm5,0),\text{ foci }(\pm7,0)$

Answer

Given that, $\text{Vertices}=(\pm5,0),\text{ foci}=(\pm7,0)$
$\therefore\ \text{a}=5\text{ and ae}=7$
$\Rightarrow\text{e}=\frac{7}{5}$
Now $\text{b}^2=\text{a}^2(\text{e}^2-1)$
$=25\Big(\frac{49}{25}-1\Big)$
$=49-25=24$
So, the equation of hyperbola is,
$\frac{\text{x}^2}{25}-\frac{\text{y}^2}{24}=1$

View full question & answer→

Question 32 Marks

Given the ellipse with equation 9x² + 25y² = 225, find the eccentricity and foci.

Answer

Given equation of ellipse, 9x² + 25y² = 225
Or $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
So, $\text{a}=5,\text{ b}=3$
Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\Rightarrow9=25(1-\text{e}^2)$
$\Rightarrow\frac{9}{25}=1-\text{e}^2$
$\Rightarrow\text{e}^2=1-\frac{9}{25}=\frac{16}{25}$
$\therefore\ \text{e}=\frac{4}{5}$
$\text{Foci}\equiv(\pm\text{ae},0)\equiv\Big(\pm\times\Big(\frac{4}{5}\Big),0\Big)\equiv(\pm4,0)$

View full question & answer→

Question 42 Marks

Find the equation of the hyperbola with:
$\text{Vertices }(0,\pm7),\text{ e}=\frac{4}{3}$

Answer

$\text{Vertices }(0,\pm7),\text{ e}=\frac{4}{3}$
$\therefore\ \text{b}=7,\text{ e}=\frac{4}{3}$
Now, $\text{a}^2=\text{b}^2(\text{e}^2-1)$
$=49\Big(\frac{16}{9}-1\Big)=\frac{343}{9}$
So, the equation of hyperbola is,
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=-1$
$\Rightarrow\frac{\text{x}^2}{\frac{343}{9}}-\frac{\text{y}^2}{49}=-1$
$\Rightarrow9\text{x}^2-7\text{y}^2+343=0$

View full question & answer→

Question 52 Marks

Find the distance between the directrices of the ellipse $\frac{\text{x}^2}{36}+\frac{\text{y}^2}{20}=1$

Answer

The equation of ellipse is $\frac{\text{x}^2}{36}+\frac{\text{y}^2}{20}=1$
$\because\text{ a}=6,\text{ b}=2\sqrt{5}$
We know that, $\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\Rightarrow20=36(1-\text{e}^2)$
$\Rightarrow\frac{5}{9}=1-\text{e}^2$
$\Rightarrow\text{e}^2=\frac{4}{9}$
$\therefore\ \text{e}=\frac{2}{3}$
Now, directrices are $\text{x}=\pm\frac{\text{a}}{\text{e}}$
$\therefore$ Distance between direcrtrix $=\frac{2\text{a}}{\text{e}}=\frac{2\times6}{\frac{2}{3}}=18$

View full question & answer→

Question 62 Marks

Find the equation of the following parabolas:
Directrix x = 0, focus at (6, 0)

Answer

We know that the distance of any point on the parabola from its focus and its directrix is same.
Given that, directrix, x = 0 and focus = (6, 0)
So, for any point P(x, y) on the parabola
Distance of P from directrix = Distance of P from focus
⇒ x² = (x - 6)² + y²
⇒ y²- 12x + 36 = 0

View full question & answer→

Question 72 Marks

Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.

Answer

Given that the circle of radius ‘a’ touches both axis. So, its centre is (a, a).

So, the equation of required circle is,
(x - a)² + (y - a)² = a²
⇒ x² - 2ax + a²+ y² - 2ay + a² = a²
⇒ x²+ y²- 2ax - 2ay + a² = 0

View full question & answer→

Question 82 Marks

If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Answer

Consider the equation of the ellipse is $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
It is given that, length of latus rectum = half of minor axis
$\Rightarrow\frac{2\text{b}^2}{\text{a}}=\text{b}$
$\Rightarrow\text{a}=2\text{b}$
Now, $\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\Rightarrow\text{b}^2=4\text{b}^2(1-\text{e}^2)$
$\Rightarrow1-\text{e}^2=\frac{1}{4}$
$\Rightarrow\text{e}^2=\frac{3}{4}$
$\therefore\ \text{e}=\frac{\sqrt{3}}{2}$

View full question & answer→

Question 92 Marks

Find the equation of the circle having (1, -2) as its centre and passing through 3x + y = 14, 2x + 5y = 18.

Answer

We get point of intersection of the lines as A(4, 2)
Now circle with centre C(1, -2) passes through A(4, 2)
$\therefore\ \text{Radius = AC}=\sqrt{(4-1)^2+(2+2)^2}$
$\text{AC}=\sqrt{9+16}=5$
So, equation of the required circle is,
(x - 1)² + (y + 2)² = 5²
⇒ x² - 2x + 1 + y² + 4y + 4 = 25
⇒ x² + y² - 2x + 4y - 20 = 0

View full question & answer→

Question 102 Marks

If the line $\text{y}=\sqrt{3}\text{x}++\text{k}$ touches the circle x² + y² = 16, then find the value of k.
[Hint: Equate perpendicular distance from the centre of the circle to its radius]

Answer

Given circle is x² + y² = 16
Center = (0, 0)
Radius r = 4
Perpendicular from the origin to the given line $\text{y}=\sqrt{3}\text{x}+\text{k}$ is equal to the radius.
$\therefore\ 4=\Bigg|\frac{0-0-\text{k}}{\sqrt{(1)^2+(\sqrt{3})^2}}\Bigg|=\Big|\frac{-\text{k}}{\sqrt{4}}\Big|$
$\Rightarrow4=\pm\frac{\text{k}}{2}$
$\Rightarrow\text{k}=\pm8$
Hence, the required values of k are $\pm8$

View full question & answer→

MATHS 2 Marks Questions

Test yourself on this topic