CBSE Class 11 Maths Important Questions - Chapter 10 Conic Sections

Very Short Answer Questions: (1 Marks)

1. Find the equation of a circle with centre (P,Q) & touching the y -  axis

(A) ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{Qy}\mathbf{+}{\mathbf{Q}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

(B)  ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{px}\mathbf{-}\mathbf{2}\mathbf{Qy}\mathbf{+}{\mathbf{Q}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

(C) ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{px}\mathbf{+}\mathbf{2}\mathbf{Qy}\mathbf{+}{\mathbf{Q}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

(D) None of these

Ans: $x^2+y^2-2px-2Qy+Q^2=0$

2. Find the equations of the directrix & the axis of the parabola $\Rightarrow \mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{y}$

(A) $\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{0}$

(B) $\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{X}\mathbf{=}\mathbf{0}$

(C) $\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{0}$

(D) None of these

Ans: $3y-4=0,x=0$

3. Find the coordinates of the foci of the ellipse $\Rightarrow {\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{100}$

(A) $\mathbf{F}\mathbf{(}\pm \mathbf{5}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{0}\mathbf{)}$

(B) $\mathbf{F}\mathbf{(}\pm \mathbf{3}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{)}$

(C) $\mathbf{F}\mathbf{(}\pm \mathbf{4}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{)}$

(D) None of these

Ans: $F(\pm 5\sqrt{3},0)$

4. Find the eccentricity of the hyperbola: $\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{6}$

1. $\mathbf{e}\mathbf{=}\sqrt{{\displaystyle \frac{\mathbf{5}}{\mathbf{2}}}}$

2. $\mathbf{e}\mathbf{=}{\displaystyle \frac{\sqrt{\mathbf{5}}}{\mathbf{2}}}$

3. $\mathbf{e}\mathbf{=}{\displaystyle \frac{\sqrt{\mathbf{2}}}{\mathbf{5}}}$

4. None of these

Ans: $e=\sqrt{{\displaystyle \frac{5}{2}}}$

5. Find the equation of a circle with centre $\mathbf{(}\mathbf{b}\mathbf{,}\mathbf{a}\mathbf{)}$ & touching $\mathbf{x}\mathbf{-}$ axis?

1. ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{bx}\mathbf{+}\mathbf{2}\mathbf{ay}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

2. ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{bx}\mathbf{-}\mathbf{2}\mathbf{ay}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

3. ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{bx}\mathbf{-}\mathbf{2}\mathbf{ay}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

4. None of these

Ans: $x^2+y^2-2bx-2ay+b^2=0$

6. Find the length of the latus rectum of $\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{18}\mathbf{?}$

1. $\mathbf{2}$ units

2. $\mathbf{3}$ units

3. $\mathbf{4}$ units

4. None of these

Ans: $4$ units

7. Find the length of the latus rectum of the parabola $\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{x}$

1. ${\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}$ units

2. ${\displaystyle \frac{\mathbf{8}}{\mathbf{3}}}$ units

3. ${\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}$ units

4. None of these

Ans: ${\displaystyle \frac{8}{3}}$ units

8. The equation ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathbf{x}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{-}\mathbf{72}\mathbf{=}\mathbf{0}$ represent a circle find its centre

1. $$\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{)}$$

2. $$\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{)}$$

3.  $$\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{4}\mathbf{)}$$

4. $$\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{4}\mathbf{)}$$

Ans: $(6,-4)$

9. Find the equation of the parabola with focus $\mathbf{F}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$ & directrix $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{4}$

1. ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{32}\mathbf{x}$

2. ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{16}\mathbf{x}$

3. ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{x}$

4. ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{16}\mathbf{x}$

Ans: $y^2=16x$

10. Find the coordinates of the foci of ${\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{8}}}\mathbf{+}{\displaystyle \frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{4}}}\mathbf{=}\mathbf{1}$

1. ${\mathbf{F}}_{\mathbf{1}}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$& ${\mathbf{F}}_{\mathbf{2}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$

2. ${\mathbf{F}}_{\mathbf{1}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$& ${\mathbf{F}}_{\mathbf{2}}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$

3. ${\mathbf{F}}_{\mathbf{1}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$& ${\mathbf{F}}_{\mathbf{2}}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$

4. None of these

Ans: $F_1(-2,0)$& $F_2(2,0)$

11. Find the coordinates of the vertices of ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$

1. $\mathbf{A}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{B}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

2. $\mathbf{A}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{B}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

3. $\mathbf{-}\mathbf{A}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{B}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

4. None of these

Ans: $A(-1,0),B(1,0)$

13. Find the eccentricity of ellipse $\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$

1. $\mathbf{e}\mathbf{=}{\displaystyle \frac{\sqrt{\mathbf{5}}}{\mathbf{3}}}$

2. $\mathbf{e}\mathbf{=}{\displaystyle \frac{\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{3}}}$

3. $\mathbf{e}\mathbf{=}{\displaystyle \frac{\sqrt{\mathbf{3}}}{\mathbf{5}}}$

4. $\mathbf{e}\mathbf{=}{\displaystyle \frac{\mathbf{3}}{\sqrt{\mathbf{5}}}}$

Ans: $e={\displaystyle \frac{\sqrt{5}}{3}}$

14. Find the length of the latus rectum of $\mathbf{9}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{36}$

1. ${\displaystyle \frac{\mathbf{1}}{\mathbf{3}}}$ units

2. ${\displaystyle \frac{\mathbf{1}}{\mathbf{5}}}$ units

3. $\mathbf{1}{\displaystyle \frac{\mathbf{1}}{\mathbf{3}}}$ units

4. ${\displaystyle \frac{\mathbf{1}}{\mathbf{6}}}$ units

Ans: $1{\displaystyle \frac{1}{3}}$ units

15. Find the length of minor axis of ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{100}$

1. $\mathbf{10}$ units

2. $\mathbf{12}$ units

3. $\mathbf{14}$ units

4. $\mathbf{8}$ units

Ans: $10$ units 

16. Find the centre of the circles ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{2}$

1. $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

2. $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$ 

3. $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}$ 

4. None of these

Ans: $(0,1)$

17. Find the radius of circles ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{2}$

1. $\sqrt{\mathbf{2}}$

2. $\mathbf{2}$ 

3. $\mathbf{2}\sqrt{\mathbf{2}}$ 

4. None of these

Ans: $\sqrt{2}$

18. Find the length of latus rectum of ${\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{22}\mathbf{y}$

1. $$\mathbf{11}$$

2. $\mathbf{-}\mathbf{22}$ 

3. $\mathbf{22}$ 

4. None of these

Ans: $22$

19. Find the length of latus rectum of $\mathbf{25}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{100}$

1. ${\displaystyle \frac{\mathbf{3}}{\mathbf{5}}}$ units

2. ${\displaystyle \frac{\mathbf{1}}{\mathbf{5}}}$ units 

3. ${\displaystyle \frac{\mathbf{8}}{\mathbf{5}}}$ units 

4. None of these

Ans: ${\displaystyle \frac{8}{5}}$ units

Long Answer Questions: (4 Marks)

1. Show that the equation ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{-}\mathbf{36}\mathbf{=}\mathbf{0}$ represents a circle, also find its centre & radius?

Ans: It is of the form $x^2+y^2+2gx+2Fy+c=0$,

Where $2g=-6,2f=4$& $c=-36$

$\therefore g=-3,f=2$& $c=-36$

Thus, center of the circle is $(-g,-f)=(3,-2)$

Radius of the circle is $\sqrt{g^2+f^2-c}=\sqrt{9+4+36}$

$=7$ units

2. Find the equation of an ellipse whose foci are $\mathbf{(}\pm \mathbf{8}\mathbf{,}\mathbf{0}\mathbf{)}$ & the eccentricity is ${\displaystyle \frac{\mathbf{1}}{\mathbf{4}}}$ ?

Ans: Let the required equation of the ellipse be ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$, where $a^2>b^2$

Let the foci be $(\pm c,0),c=8$

And $e={\displaystyle \frac{c}{a}}$

$\Rightarrow a={\displaystyle \frac{c}{e}}$

$\Rightarrow a={\displaystyle \frac{8}{{\displaystyle \frac{1}{4}}}}$

$\Rightarrow a=32$

As $c^2=a^2-b^2$

$\Rightarrow b^2=a^2-c^2$

$\Rightarrow b^2=1024-64$

$\Rightarrow b^2=960$

$\therefore a^2=1024$

& $b^2=960$

Therefore the equation is ${\displaystyle \frac{x^2}{1024}}+{\displaystyle \frac{y^2}{960}}=1$

3. Find the equation of an ellipse whose vertices are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{10}\mathbf{)}$& $\mathbf{e}\mathbf{=}{\displaystyle \frac{\mathbf{4}}{\mathbf{5}}}$

Ans: Let equation be ${\displaystyle \frac{x^2}{b^2}}+{\displaystyle \frac{y^2}{a^2}}=1$

Vertices are $(0,\pm a),a=10$

Let $c^2=a^2-b^2$

So, $e={\displaystyle \frac{c}{a}}$

$\Rightarrow c=ae$

$\Rightarrow c=10\times {\displaystyle \frac{4}{5}}$

$\Rightarrow c=8$

Now, $c^2=a^2-b^2$

$\Rightarrow b^2=\left(a^2-c^2\right)$

$\Rightarrow b^2=100-64$

$\Rightarrow b^2=36$

So, $a^2=(10{)}^2=100$ & $b^2=36$

Therefore the equation is ${\displaystyle \frac{x^2}{36}}+{\displaystyle \frac{y^2}{100}}=1$

4. Find the equation of hyperbola whose length of latus rectum is $\mathbf{36}$ & foci are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{12}\mathbf{)}$

Ans: It is clear that $c=12$.

Length of latus rectum $=36$

$\Rightarrow {\displaystyle \frac{2b^2}{a}}=36$

$\Rightarrow b^2=18a$

Now, $c^2=a^2+b^2$

$\Rightarrow a^2=c^2-b^2$

$\Rightarrow a^2=144-18a$

$\Rightarrow a^2+18a-144=0$

$\Rightarrow (a+24)(a-6)=0$

$\Rightarrow a=6$ because $a$ is non-negative.

Thus $a^2=6^2=36$ & $b^2=108$

Therefore, ${\displaystyle \frac{x^2}{36}}+{\displaystyle \frac{y^2}{108}}=1$

5. Find the equation of a circle drawn on the diagonal of the rectangle as its diameter, whose sides are $\mathbf{x}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{3}$ & $\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}$

Ans: Let $\text{ABCD}$ be the given rectangle and $AD=x=-3,BC=x=6,AB=y=-1$ & $CD=y=-3$.

Then $A(-3,-1)$ and $C(6,3)$.

The equation of the circle with $AC$ as diameter is:

$(x+3)(x-6)+(y+1)(y-3)=0$

$\Rightarrow x^2+y^2-3x-2y-21=0$

6. Find the coordinates of the focus & vertex, the equations of the directrix & the axis & length of latus rectum of the parabola ${\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{y}$

Ans: $x^2=-8y$ & $x^2=-4ay$

So, $4a=8$

$\Rightarrow a=2$

So it is downward parabola.

Foci is $F(0,-a)$ i.e. $F(0,-2)$.

Vertex is $O(0,0)$.

So, $y=a=2$.

Its axis is $y-$ axis, whose equation is given by $x=0$.

Length of latus rectum$=4a$ units.

$=4\times 2$ units

$=8$ units

7. Show that the equation $\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{24}\mathbf{x}\mathbf{-}\mathbf{36}\mathbf{y}\mathbf{-}\mathbf{18}\mathbf{=}\mathbf{0}$ represents a circle. Also find its centre & radius.

Ans: $6x^2+6y^2+24x-36y-18=0$

So, $x^2+y^2+4x-6y+3=0$

Where, $2g=4,2f=-6$ & $c=3$

$\therefore g=2,f=-3$ & $c=3$

Thus, centre of circle is $(-g,-f)=(-2,3)$

Radius of circle $=\sqrt{4+9+9}=\sqrt{20}$

$=2\sqrt{5}$ units

8. Find the equation of the parabola with focus at $\mathbf{F}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}$ & directrix is $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{5}$

Ans: $F(5,0)$ lies on the right hand side of origin.

Thus, it is a right hand parabola.

Let the required equation be

$y^2=4ax$ & $a=5$

Hence, $y^2=20x$

9. Find the equation of the hyperbola with center at the origin, length of the transverse axis $\mathbf{6}$ & one focus at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{4}\mathbf{)}$

Ans: Let its equation be ${\displaystyle \frac{y^2}{a^2}}-{\displaystyle \frac{x^2}{b^2}}=1$

It is clear that $c=4$.

Length of the transverse axis $=6$

$\Rightarrow 2a=6$

$\Rightarrow a=3$

And, $c^2=\left(a^2+b^2\right)$

$\Rightarrow b^2=c^2-a^2$

$\Rightarrow b^2=16-9$

$\Rightarrow b^2=7$

Thus, $a^2=9$ & $b^2=7$

Hence, equation is ${\displaystyle \frac{x^2}{9}}-{\displaystyle \frac{x^2}{7}}=1$

10. Find the equation of an ellipse whose vertices are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{13}\mathbf{)}$ & the foci are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{5}\mathbf{)}$

Ans: Let the equation be ${\displaystyle \frac{x^2}{b^2}}+{\displaystyle \frac{y^2}{a^2}}=1$

& $a=13$

Let foci be $(0,\pm c)$,

$\Rightarrow c=5$

$\therefore b^2=a^2-c^2$

$\Rightarrow b^2=169-25$

$\Rightarrow b^2=144$

$\Rightarrow a^2=169$

And $b^2=144$

Thus, equation is ${\displaystyle \frac{x^2}{144}}+{\displaystyle \frac{y^2}{169}}=1$

11. Find the equation of the ellipse whose foci are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{3}\mathbf{)}$ & length of whose major axis is $\mathbf{10}$

Ans:Let the required equation be ${\displaystyle \frac{x^2}{b^2}}+{\displaystyle \frac{y^2}{a^2}}=1$

Let $c^2=a^2-b^2$

Foci are $(0,\pm c)$ & $c=3$

$a=$length of the semi- major axis i.e. ${\displaystyle \frac{1}{2}}\times 10=5$

So, $c^2=a^2-b^2$

$\Rightarrow b^2=a^2-c^2$

$\Rightarrow b^2=25-3$

$\Rightarrow b^2=16$

Thus, $a^2=25$ & $b^2=16$

So, the required equation is ${\displaystyle \frac{x^2}{16}}+{\displaystyle \frac{y^2}{25}}=1$.

12. Find the equation of the hyperbola with centre at the origin, length of the transverse axis $\mathbf{8}$ & one focus at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{6}\mathbf{)}$

Ans: Let its equation by ${\displaystyle \frac{y^2}{a^2}}-{\displaystyle \frac{x^2}{b^2}}=1$

It is clear that $c=6$ 

And the length of the transverse axis $=8$

$\Rightarrow 2a=8$

$\Rightarrow a=4$

And, $c^2=a^2+b^2$

$\Rightarrow b^2=c^2-a^2$

$\Rightarrow 36-16=20$

So, $a^2=16$ & $b^2=20$

So, the required equation is ${\displaystyle \frac{y^2}{16}}-{\displaystyle \frac{x^2}{20}}=1$

13. Find the equation of the hyperbola whose foci are at $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{B}\mathbf{)}$ & the length of whose conjugate axis is $\mathbf{2}\sqrt{\mathbf{11}}$

Ans: Let equation be ${\displaystyle \frac{y^2}{a^2}}-{\displaystyle \frac{x^2}{b^2}}=1$

Let foci be $(0,\pm c)$

$\therefore c=8$

Length of conjugate axis $=2\sqrt{11}$

$\Rightarrow 2b=2\sqrt{11}$

$\Rightarrow b=\sqrt{11}$

$\Rightarrow b^2=11$

And, $c^2=\left(a^2+b^2\right)$

$\Rightarrow a^2=\left(c^2-b^2\right)$

$\Rightarrow a^2=64-11$

$\Rightarrow a^2=53$

So, the required equation is ${\displaystyle \frac{y^2}{53}}-{\displaystyle \frac{x^2}{11}}=1$

14. Find the equation of the hyperbola whose vertices are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{3}\mathbf{)}$ & foci are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{8}\mathbf{)}$

Ans: Vertices are $(0\pm a)$

It is given that the vertices are $(0\pm 3)$

$\therefore a=3$

Let foci be $(0,\pm c)$

It is given that the foci are $(0,\pm 8)$

$\therefore c=8$

And $b^2=\left(c^2-a^2\right)$

$\Rightarrow b^2=8^2-3^2$

$\Rightarrow b^2=64-9$

$\Rightarrow b^2=55$

Now, $a^2=3^2=9$ & $b^2=55$.

So, the required equation is ${\displaystyle \frac{y^2}{9}}-{\displaystyle \frac{x^2}{55}}=1$

15. Find the equation of the ellipse for which $\mathbf{e}\mathbf{=}{\displaystyle \frac{\mathbf{4}}{\mathbf{5}}}$ & whose vertices are $\mathbf{(}\mathbf{0.}\pm \mathbf{10}\mathbf{)}$.

Ans: Vertices are $(0,\pm a)$ 

So, $a=10$

Let $c^2=\left(a^2-b^2\right)$

$\Rightarrow e={\displaystyle \frac{c}{a}}$

$\Rightarrow c=ae$

$\Rightarrow c=\left[10\times {\displaystyle \frac{4}{5}}\right]$

$\Rightarrow c=8$

And, $c^2=\left(a^2-b^2\right)$

$\Rightarrow b^2=\left(a^2-c^2\right)$

$\Rightarrow b^2=(100-64)$

$\Rightarrow b^2=36$

$\therefore a^2=(10{)}^2=100$ & $b^2=36$

So, the required equation is ${\displaystyle \frac{x^2}{36}}+{\displaystyle \frac{y^2}{100}}=1$

16. Find the equation of the parabola with vertex at the origin & $\mathbf{\text{y}}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0}$ as its directrix. Also, find its focus

Ans: Let the vertex of the parabola be $O(0,0)$.

$y+5=0$

$\Rightarrow y=-5$

The directrix is a line parallel to the $x-$axis at a distance of $5$ units below the $x-$axis. Thus, the focus is $F(0,5)$.

So, the equation of the parabola is $x^2=4ay$ where $a=5$ i.e. $x^2=20y$.

17. Find the equation of a circle, the end points of one of whose diameters are $\mathbf{A}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$& $\mathbf{B}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{)}$

Ans: Let the end points of one of whose diameters are $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

So, $x_1=2,y_1=-3$ & $x_2=-3,y_2=5$.

$\therefore$ The required equation of the circle is $(x-2)(x+3)+(y+3)(y-5)=0$

$\Rightarrow x^2+y^2+x-2y-21=0$

18. Find the equation of ellipse whose vertices are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{13}\mathbf{)}$ & the foci are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{5}\mathbf{)}$

Ans: Let the required equation be ${\displaystyle \frac{x^2}{b^2}}+{\displaystyle \frac{y^2}{a^2}}=1$.

Its vertices are $(0\pm a)$.

So, $a=13$

Let its foci be $(0\pm c)$ then $c=5$.

$\therefore b^2=a^2-c^2$

$\Rightarrow b^2=169-25$

$\Rightarrow b^2=144$

Thus $b^2=144$ and $a^2=169$.

So, the required equation is ${\displaystyle \frac{x^2}{144}}+{\displaystyle \frac{y^2}{169}}=1$.

19. Find the equation of the hyperbola whose foci are $\mathbf{(}\pm \mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}$ & the transverse axis is of length $\mathbf{8}$.

Ans: Let the required equation be ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$

Length of transverse axis$=2a$.

$\therefore 2a=8$

$\Rightarrow a=4$

$\Rightarrow a^2=16$

Let its foci be $(\pm c,0)$.

So, $c=5$.

$\therefore b^2=\left(c^2-a^2\right)$

$\Rightarrow b^2=5^2-4^2$

$\Rightarrow b^2=9$

Thus $a^2=16$ and $b^2=9$

Therefore, the required equation is ${\displaystyle \frac{x^2}{16}}-{\displaystyle \frac{y^2}{9}}=1$.

20. Find the equation of a circle, the end points of one of whose diameters are $\mathbf{A}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{)}$& $\mathbf{B}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$

Ans: Let the equation be $\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

So $x_1=-3,y_1=2$ and $x_2=5,y_2=-3$.

$\Rightarrow (x+3)(x-5)+(y-2)(y+3)=0$

$\Rightarrow x^2-2x-15+y^2+y-6=0$

$\Rightarrow x^2+y^2-2x+y-21=0$

21. If eccentricity is ${\displaystyle \frac{\mathbf{1}}{\mathbf{5}}}$ & foci are $\mathbf{(}\pm \mathbf{7}\mathbf{,}\mathbf{0}\mathbf{)}$ find the equation of an ellipse.

Ans: Let the required equation of the ellipse be ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$.

Let its foci be $(\pm c,0)$.

So, $c=7$.

And $e={\displaystyle \frac{c}{a}}$

$\Rightarrow a={\displaystyle \frac{c}{e}}$

$\Rightarrow a={\displaystyle \frac{7}{{\displaystyle \frac{1}{5}}}}$

$\Rightarrow a=35$

Also $c^2=\left(a^2-b^2\right)$

$\Rightarrow b^2=a^2-c^2$

$\Rightarrow b^2=(35{)}^2-49$

$\Rightarrow b^2=1225-49$

$\Rightarrow b^2=1176$

$\therefore a^2=1225$ and $b^2=1176$.

So, the required equation is ${\displaystyle \frac{x^2}{1225}}+{\displaystyle \frac{y^2}{1176}}=1$

22. Find the equation of the hyperbola whose foci are $\mathbf{(}\pm \mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}$ & the transverse axis is of length

Ans: Let the required equation be ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$

Length of its transverse axis $=2a$

$\therefore 2a=8$

$\Rightarrow a=4$

$\Rightarrow a^2=16$

Let its foci be $(\pm c,0)$

So, $c=5$.

$\therefore b^2=c^2-a^2$

$\Rightarrow b^2=25-16$

$\Rightarrow b^2=9$

So, the required equation is ${\displaystyle \frac{x^2}{16}}-{\displaystyle \frac{y^2}{9}}=1$

23. Find the length of axes & coordinates of the vertices of the hyperbola ${\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{49}}}\mathbf{-}{\displaystyle \frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{64}}}\mathbf{=}\mathbf{1}$

Ans: The equation of the given hyperbola is ${\displaystyle \frac{x^2}{49}}-{\displaystyle \frac{y^2}{64}}=1$

Compare the given equation with ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$.

$\Rightarrow a^2=49$ and $b^2=64$.

$\therefore c^2=\left(a^2+b^2\right)$

$\Rightarrow c^2=49+64$

$\Rightarrow c^2=113$

Length of transverse axis is $2a=2\times 7=14$ units

Length of conjugate axis is $2b=2\times 8=16$ units

The coordinates of the vertices are $A(-a,0)$ and $B(a,0)$ i.e. $A(-7,0)$ and $B(7,0)$.

24. Find the lengths of axes & length of latus rectum of the hyperbola, ${\displaystyle \frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}}\mathbf{-}{\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}}\mathbf{=}\mathbf{1}$

Ans: The given equation is ${\displaystyle \frac{y^2}{9}}-{\displaystyle \frac{x^2}{16}}=1$.

Compare the given equation with ${\displaystyle \frac{y^2}{a^2}}-{\displaystyle \frac{x^2}{b^2}}=1$.

$\Rightarrow a^2=9$ & $b^2=16$

Length of transverse axis is $2a=2\times 3=6$ units

Length of conjugate axis is $2b=2\times 4=8$ units

The coordinates of the vertices are $A(0,-a)$ and $B(0,a)$ i.e. $A(0,-3)$ and $B(0,3)$.

25. Find the eccentricity of the hyperbola of ${\displaystyle \frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}}\mathbf{-}{\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}}\mathbf{=}\mathbf{1}$

Ans: Here, $a=3$ and $b=4$

And $c^2=a^2+b^2$

$\Rightarrow c^2=9+16$

$\Rightarrow c^2=25$

Thus, $c=5$

$\Rightarrow e={\displaystyle \frac{c}{a}}$

$\Rightarrow e={\displaystyle \frac{5}{3}}$

26. Find the equation of the ellipse, the ends of whose major axis are $\mathbf{(}\pm \mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}$ & at the ends of whose minor axis are $\mathbf{(}\mathbf{0}\mathbf{,}\pm \mathbf{4}\mathbf{)}$

Ans: Let the required equation be ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$

Its vertices are $(\pm a,0)$.

So, $a=3$.

Ends of minor axis are $C(0,-4)$ and $D(0,4)$.

$\therefore CD=8$ i.e. length of the minor axis$=8$ units

$\Rightarrow 2b=8$

$\Rightarrow b=4$

$\therefore a=3$ and $b=4$.

Therefore, the required equation is ${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{16}}=1$.

27. Find the equation of the parabola with focus at $\mathbf{F}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$ & directrix $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}$

Ans: Focus $F(4,0)$ lies on the axis hand side of the origin.

Thus, it is a right handed parabola.

Let the required equation be $y^2=4ax$.

Then, $a=4$.

So, the required equation is $y^2=16x$.

28. If $\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{x}$ is a chord of the circle ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{10}\mathbf{x}\mathbf{=}\mathbf{0}$, find the equation of the circle with this chord as a diameter

Ans: $y=2x$ and $x^2+y^2-10x=0$

Put $y=2x$ in $x^2+y^2-10x=0$.

$\Rightarrow 5x^2-10x=0$

$\Rightarrow 5x(x-2)=0$

$\Rightarrow x=0$ or $x=2$

Now, $x=0\Rightarrow y=0$ and $x=2\Rightarrow y=4$.

$\therefore$ The points of intersection of the given chord and the given circle are$A(0,0)$ and $B(2,4)$.

$\therefore$ The required equation of the circle with $\text{AB}$ as diameter is $(x-0)(x-2)+(y-0)(y-4)=0$

$\Rightarrow x^2+y^2-2x-4y=0$

Very Long Answer Questions: (6 Marks)

1. Find the length of major & minor axis- coordinates of vertices & the foci, the eccentricity & length of latus rectum of the ellipse $\mathbf{16}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{16}$

Ans: $16x^2+y^2=16$

Divide by $16$,

$\Rightarrow x^2+{\displaystyle \frac{y^2}{16}}=1$

i.e. $b^2=1$ and $a^2=16$

So, $b=1$ & $a=4$.

And $c=\sqrt{a^2-b^2}$

$\Rightarrow c=\sqrt{16-1}$

$\Rightarrow c=\sqrt{15}$

So, $a=4,b=1$ & $c=\sqrt{15}$.

(i) Length of major axis$=2a=2\times 4=8$ units

Length of minor axis$=2b=2\times 1=2$ units

(ii) Coordinates of the vertices are $A(-a,0)$ & $B(a,0)$ i.e. $A(-4,0)$ & $B(4,0)$

(iii) Coordinates of foci are $F_1(-c,0)$ & $F_2(c,0)$ i.e. $F_1(-\sqrt{15},0)$ & $F_2(\sqrt{15},0)$

(iv) Eccentricity, $e={\displaystyle \frac{c}{a}}={\displaystyle \frac{\sqrt{15}}{4}}$

(v) Length of latus rectum$={\displaystyle \frac{2b^2}{a}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}$ units

2. Find the lengths of the axis, the coordinates of the vertices & the foci the eccentricity & length of the latus rectum of the hyperbola $\mathbf{25}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{9}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{225}$

Ans: $$25x^2-9y^2=225$$

$$\Rightarrow {\displaystyle \frac{x^2}{9}}-{\displaystyle \frac{y^2}{25}}=1$$

Now, $a^2=9$ & $b^2=25$

And $c=\sqrt{a^2+b^2}$

$\Rightarrow c=\sqrt{9+25}$

$\Rightarrow c=\sqrt{34}$

(i) Length of transverse axis $=2a=2\times 3=6$ units

Length of conjugate axis $=2b=2\times 5=10$ units

(ii) The coordinates of vertices are $A(-a,0)$ & $B(a,0)$ i.e. $A(-3,0)$ & $B(3,0)$

(iii) The coordinates of foci are $F_1(-c,0)$ & $F_2(c,0)$ i.e. $F_1(-\sqrt{34},0)$ & $F_2(\sqrt{34},0)$

(iv) Eccentricity, $e={\displaystyle \frac{c}{a}}={\displaystyle \frac{\sqrt{34}}{3}}$

(v) Length of the latus rectum $={\displaystyle \frac{2b^2}{a}}={\displaystyle \frac{50}{3}}$ units

3. A man running in a race course notes that the sum of the distances of the two flag posts from him is always $\mathbf{12}\mathbf{\text{ m}}$& the distance between the flag posts is $\mathbf{10}\mathbf{\text{ m}}$. Find the equation of the path traced by the man.

Ans: Ellipse is the locus of a point that moves in such a way that the sum of its distances from two fixed points is constant.

Thus, the path is an ellipse.

Let the equation of the ellipse be ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$.

Where $b^2=a^2\left(1-c^2\right)$

It is clear that $2a=12$ & $2ae=10$

$\Rightarrow a=b$ and $e={\displaystyle \frac{5}{6}}$

$\Rightarrow b^2=a^2\left(1-e^2\right)$

$\Rightarrow b^2=36\left(1-{\displaystyle \frac{25}{36}}\right)$

$\Rightarrow b^2=11$

So, the required equation is ${\displaystyle \frac{x^2}{36}}+{\displaystyle \frac{y^2}{11}}=1$.

4. An equilateral triangle is inscribed in the parabola ${\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{ax}$ so that one angular point of the triangle is at the vertex of the parabola. Find the length of each side of the triangle.

Ans: Let $OPQ$ be an equilateral triangle inscribed in the parabola $y^2=4ax$ where $O(0,0)$ is the vertex so that $\mathrm{\angle }POM=\mathrm{\angle }QOM={30}^{\circ }$.

Let $OP=OQ=r$.

And $P=(r\cos {30}^{\circ },r\sin {30}^{\circ })$

$\Rightarrow P=\left({\displaystyle \frac{r\sqrt{3}}{2}},{\displaystyle \frac{r}{2}}\right)$