2 Marks Questions

Question 12 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
$\Rightarrow x^2 = (x - 6)^2 + y^2$
$\Rightarrow y^{2 }- 12x + 36 = 0$

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Question 22 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$,$
$(x - a)^2 + (y - a)^2 = a^2$
$\Rightarrow x^2 - 2ax + a^2+ y^2 - 2ay + a^2 = a^2$
$\Rightarrow x^2+ y^2- 2ax - 2ay + a^2 = 0$

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Question 32 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,
$\frac{\text{x}^2}{25}-\frac{\text{y}^2}{24}=1$

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Question 42 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, $\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\Rightarrow\text{b}^2=4\text{b}^2(1-\text{e}^2)$
$\Rightarrow1-\text{e}^2=\frac{1}{4}$
$\Rightarrow\text{e}^2=\frac{3}{4}$
$\therefore\ \text{e}=\frac{\sqrt{3}}{2}$

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Question 52 Marks

Given the ellipse with equation $9x^2 + 25y^2 = 225$, find the eccentricity and foci.

Answer

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

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Question 62 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,
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=-1$
$\Rightarrow\frac{\text{x}^2}{\frac{343}{9}}-\frac{\text{y}^2}{49}=-1$
$\Rightarrow9\text{x}^2-7\text{y}^2+343=0$

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Question 72 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 $A(4, 2)$
$\therefore$ Radius $= AC =\sqrt{(4-1)^2+(2+2)^2}$
$\text{AC}=\sqrt{9+16}=5$
So, equation of the required circle is,
$(x - 1)^2 + (y + 2)^2 = 5^2$
$\Rightarrow x^2 - 2x + 1 + y^2 + 4y + 4 = 25$
$\Rightarrow x^2 + y^2 - 2x + 4y - 20 = 0$

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Question 82 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 $=\frac{2\text{a}}{\text{e}}=\frac{2\times6}{\frac{2}{3}}=18$

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Question 92 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 $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$

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Question 102 Marks

If the line $\text{y}=\sqrt{3}\text{x}+\text{k}$ touches the circle $x^2 + y^2 = 16,$ then find the value of $k$.
$[$Hint: Equate perpendicular distance from the centre of the circle to its radius$]$

Answer

Given circle is $x^2 + y^2 = 16$
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 $\pm\ 8$

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