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

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

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

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

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

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

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

(D) None of these

Ans: $3y - 4 = 0,{\text{ }}x = 0$

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

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

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

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

(D) None of these

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

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

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

2. $\mathbf{e = \dfrac{{\sqrt 5 }}{2}}$

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

4. None of these

Ans: $e = \sqrt {\dfrac{5}{2}} $

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

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

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

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

4. None of these

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

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

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

1. $\mathbf{\dfrac{4}{3}}$ units

2. $\mathbf{\dfrac{8}{3}}$ units

3. $\mathbf{\dfrac{2}{3}}$ units

4. None of these

Ans: $\dfrac{8}{3}$ units

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

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

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

3.  \[\mathbf{(6,4)}\]

4. \[\mathbf{( - 6,4)}\]

Ans: $(6, - 4)$

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

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

2. $\mathbf{{y^2} =  - 16x}$

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

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

Ans: ${y^2} = 16x$

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

1. $\mathbf{{F_1}(2,0)}$& $\mathbf{{F_2}( - 2,0)}$

2. $\mathbf{{F_1}( - 2,0)}$& $\mathbf{{F_2}(2,0)}$

3. $\mathbf{{F_1}( - 2,0)}$& $\mathbf{{F_2}( - 2,0)}$

4. None of these

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

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

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

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

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

4. None of these

Ans: $A( - 1,0),{\text{ }}B(1,0)$

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

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

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

3. $\mathbf{e = \dfrac{{\sqrt 3 }}{5}}$

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

Ans: $e = \dfrac{{\sqrt 5 }}{3}$

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

1. $\mathbf{\dfrac{1}{3}}$ units

2. $\mathbf{\dfrac{1}{5}}$ units

3. $\mathbf{1\dfrac{1}{3}}$ units

4. $\mathbf{\dfrac{1}{6}}$ units

Ans: $1\dfrac{1}{3}$ units

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

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 $\mathbf{{x^2} + {(y - 1)^2} = 2}$

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

2. $\mathbf{(0,1)}$ 

3. $\mathbf{(1,2)}$ 

4. None of these

Ans: $(0,1)$

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

1. $\mathbf{\sqrt 2 }$

2. $\mathbf{2}$ 

3. $\mathbf{2\sqrt 2} $ 

4. None of these

Ans: $\sqrt 2 $

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

1. \[\mathbf{11}\]

2. $ \mathbf{- 22}$ 

3. $\mathbf{22}$ 

4. None of these

Ans: $22$

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

1. $\mathbf{\dfrac{3}{5}}$ units

2. $\mathbf{\dfrac{1}{5}}$ units 

3. $\mathbf{\dfrac{8}{5}}$ units 

4. None of these

Ans: $\dfrac{8}{5}$ units

Long Answer Questions: (4 Marks)

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

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

Where $2g =  - 6,{\text{ }}2f = 4$& $c =  - 36$

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

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 $\mathbf{( \pm 8,0)}$ & the eccentricity is $\mathbf{\dfrac{1}{4}}$ ?

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

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

And $e = \dfrac{c}{a}$

$ \Rightarrow a = \dfrac{c}{e}$

$ \Rightarrow a = \dfrac{8}{{\dfrac{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 $\dfrac{{{x^2}}}{{1024}} + \dfrac{{{y^2}}}{{960}} = 1$

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

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

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

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

So, $e = \dfrac{c}{a}$

$ \Rightarrow c = ae$

$ \Rightarrow c = 10 \times \dfrac{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 $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{100}} = 1$

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

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

Length of latus rectum $ = 36$

$ \Rightarrow \dfrac{{2{b^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, $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{108}} = 1$

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

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

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 $\mathbf{{x^2} =  - 8y}$

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

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

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

Where, $2g = 4,{\text{ }}2f =  - 6$ & $c = 3$

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

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 $\mathbf{F(5,0)}$ & directrix is $\mathbf{x =  - 5}$

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 $\mathbf{(0,4)}$

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

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 $\dfrac{{{x^2}}}{{9}} - \dfrac{{{x^2}}}{{7}} = 1$

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

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

& $a = 13$

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 $\dfrac{{{x^2}}}{{144}} + \dfrac{{{y^2}}}{{169}} = 1$

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

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

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

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

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

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 $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1$.

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

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

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 $\dfrac{{{y^2}}}{{16}} - \dfrac{{{x^2}}}{{20}} = 1$

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

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

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 $\dfrac{{{y^2}}}{{53}} - \dfrac{{{x^2}}}{{11}} = 1$

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

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 $\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{55}} = 1$

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

Ans: Vertices are $(0, \pm a)$ 

So, $a = 10$

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

$ \Rightarrow e = \dfrac{c}{a}$

$ \Rightarrow c = ae$

$ \Rightarrow c = \left[ {10 \times \dfrac{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 $\dfrac{{{x^2}}}{{36}} + \dfrac{{{y^2}}}{{100}} = 1$

16. Find the equation of the parabola with vertex at the origin & $\mathbf{{\text{y}} + 5 = 0}$ 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 $\mathbf{A(2, - 3)}$& $\mathbf{B( - 3,5)}$

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, ${x_1} = 2,{\text{ }}{y_1} =  - 3$ & ${x_2} =  - 3,{\text{ }}{y_2} = 5$.

$\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 $\mathbf{(0, \pm 13)}$ & the foci are $\mathbf{(0, \pm 5)}$

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

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 $\dfrac{{{x^2}}}{{144}} + \dfrac{{{y^2}}}{{169}} = 1$.

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

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

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 $\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1$.

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

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 ${x_1} =  - 3,{\text{ }}{y_1} = 2$ and ${x_2} = 5,{\text{ }}{y_2} =  - 3$.

$ \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 $\mathbf{\dfrac{1}{5}}$ & foci are $\mathbf{( \pm 7,0)}$ find the equation of an ellipse.

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

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

So, $c = 7$.

And $e = \dfrac{c}{a}$

$ \Rightarrow a = \dfrac{c}{e}$

$ \Rightarrow a = \dfrac{7}{{\dfrac{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 $\dfrac{{{x^2}}}{{1225}} + \dfrac{{{y^2}}}{{1176}} = 1$

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

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

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 $\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1$

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

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

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

$ \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, $\mathbf{\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{16}} = 1}$

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

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

$ \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 $\mathbf{\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{16}} = 1}$

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 = \dfrac{c}{a}$

$ \Rightarrow e = \dfrac{5}{3}$

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

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

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 $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{16}} = 1$.

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

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 $\mathbf{y = 2x}$ is a chord of the circle $\mathbf{{x^2} + {y^2} - 10x = 0}$, 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 5{x^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 $\mathbf{16{x^2} + {y^2} = 16}$

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

Divide by $16$,

$ \Rightarrow {x^2} + \dfrac{{{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, $a = 4,{\text{ }}b = 1$ & $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, $e = \dfrac{c}{a} = \dfrac{{\sqrt {15} }}{4}$

(v) Length of latus rectum$ = \dfrac{{2{b^2}}}{a} = \dfrac{2}{4} = \dfrac{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 $\mathbf{25{x^2} - 9{y^2} = 225}$

Ans: \[25{x^2} - 9{y^2} = 225\]

\[ \Rightarrow \dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{25}} = 1\]

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, $e = \dfrac{c}{a} = \dfrac{{\sqrt {34} }}{3}$

(v) Length of the latus rectum $ = \dfrac{{2{b^2}}}{a} = \dfrac{{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 $\mathbf{12{\text{ m}}}$& the distance between the flag posts is $\mathbf{10{\text{ m}}}$. 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 $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.

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

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

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

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

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

$ \Rightarrow {b^2} = 11$

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

4. An equilateral triangle is inscribed in the parabola $\mathbf{{y^2} = 4ax}$ 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 $\angle POM = \angle QOM = {30^\circ }$.

Let $OP = OQ = r$.

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

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

$P$ lies on parabola. 

$ \Rightarrow \dfrac{{{r^2}}}{4} = 4a\left( {\dfrac{{r\sqrt 3 }}{2}} \right)$

$ \Rightarrow r = 8a\sqrt 3 $

Hence, the length of each side of the triangle is $8a\sqrt 3 $ units.

5. Find the equation of the hyperbola whose foci are at $\mathbf{(0, \pm \sqrt {10} )}$ & which passes through the points $\mathbf{(2,3)}$

Ans: Let equation be \[\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1{\text{      }}...{\text{(}}i)\]

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

But the foci are $(0, \pm \sqrt {10} )$.

$\therefore c = \sqrt {10} $

$ \Rightarrow {c^2} = 10$

$ \Rightarrow \left( {{a^2} + {b^2}} \right) = 10{\text{    }}...{\text{(}}ii)$

As \[(i)\] passes through $(2,3)$,

$ \Rightarrow \dfrac{9}{{{a^2}}} - \dfrac{4}{{{b^2}}} = 1$

$ \Rightarrow \dfrac{9}{{{a^2}}} - \dfrac{4}{{\left( {10 - {a^2}} \right)}} = 1$

$ \Rightarrow 9\left( {10 - {a^2}} \right) - 4{a^2} = {a^2}\left( {10 - {a^2}} \right)$

$ \Rightarrow {a^4} - 23{a^2} + 90 = 0$

$ \Rightarrow \left( {{a^2} - 18} \right)\left( {{a^2} - 5} \right) = 0$

$ \Rightarrow {a^2} = 5$

$\because {a^2} = 18 \Rightarrow {b^2} =  - 8$, which is impossible

So, ${a^2} = 5$ and ${b^2} = 5$.

So, the required equation is $\dfrac{{{y^2}}}{5} - \dfrac{{{x^2}}}{5} = 1$,

i.e. ${y^2} - {x^2} = 5$.

6. Find the equation of the curve formed by the set of all these points the sum of whose distance from the points $\mathbf{A(4,0,0)}$ & $\mathbf{B( - 4,0,0)}$ is $\mathbf{10}$ units.

Ans: Let $P(x,y,z)$ be an arbitrary point on the given curve.

So, $PA + PB = 10$

$ \Rightarrow \sqrt {{{(x - 4)}^2} + {y^2} + {z^2}}  + \sqrt {{{(x + 4)}^2} + {y^2} + {z^2}}  = 10$

$ = \sqrt {{{(x + 4)}^2} + {y^2} + {z^2}}  = 10 - \sqrt {{{(x - 4)}^2} + {y^2} + {z^2}} $

Squaring both sides:

$ \Rightarrow {(x + 4)^2} + {y^2} + {z^2} = 100 + {(x - 4)^2} + {y^2} + {z^2} - 20\sqrt {{{(x - 4)}^2} + {y^2} + {z^2}} $

$ \Rightarrow 16x = 100 - 20\sqrt {{{(x - 4)}^2} + {y^2} + {z^2}} $

$ \Rightarrow 5\sqrt {{{(x - 4)}^2} + {y^2} + {z^2}}  = 25 - 4x$

$ \Rightarrow 25\left[ {{{(x - 4)}^2} + {y^2} + {z^2}} \right] = 625 + 16{x^2} - 200x$

$ \Rightarrow 9{x^2} + 25{y^2} + 25{z^2} - 225 = 0$

So, the required equation of the curve is $9{x^2} + 25{x^2} + 25{z^2} - 225 = 0$.

7. Find the equation of the ellipse with centre at the origin, major axis on the $\mathbf{y - } $axis & passing through the points $\mathbf{(3,2)}$ & $\mathbf{(1,6)}$

Ans: Let the required equation be $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1{\text{         }}...(i)$

Since $(3,2)$ lies on $(i)$, $\dfrac{9}{{{b^2}}} + \dfrac{4}{{{a^2}}} = 1{\text{             }} \ldots (ii)$

As $(1,6)$ lies on $(i)$, $\dfrac{1}{{{b^2}}} + \dfrac{{36}}{{{a^2}}} = 1{\text{                }} \ldots (iii)$

Put $\dfrac{1}{{{b^2}}} = u$ & $\dfrac{1}{{{a^2}}} = v$.

$ \Rightarrow 9u + 4v = 1{\text{            }} \ldots (iv)$ and $u + 36v = 1{\text{             }} \ldots (v)$

Multiply $(v)$ by $9$ & subtract $(iv)$ from it.

$ \Rightarrow 320v = 8$

$ \Rightarrow v = \dfrac{8}{{320}}$

$ \Rightarrow v = \dfrac{1}{{40}}$

$ \Rightarrow \dfrac{1}{{{a^2}}} = \dfrac{1}{{40}}$

$ \Rightarrow {a^2} = 40$

Put $v = \dfrac{1}{{40}}$ in $(v)$

$ \Rightarrow u + \left[ {36 \times \dfrac{1}{{40}}} \right] = 1$

$ \Rightarrow u = \left[ {1 - \dfrac{9}{{10}}} \right] = \dfrac{1}{{10}}$

$ \Rightarrow \dfrac{1}{{{b^2}}} = \dfrac{1}{{10}}$

$ \Rightarrow {b^2} = 10$

So, ${b^2} = 10$ and ${a^2} = 40$

So, the required equation is $\dfrac{{{x^2}}}{{10}} + \dfrac{{{y^2}}}{{40}} = 1$.

Benefits of Class 11 Maths Chapter 10: Conic Sections - Important Questions

• The important questions for Class 11 Maths Chapter 10: Conic Sections cover all the essential topics like circles, parabolas, ellipses, and hyperbolas. They focus on standard equations, properties, and problem-solving techniques, ensuring a complete understanding of the chapter.

• These questions are curated based on the latest CBSE syllabus and past exam trends. They help students prepare for commonly asked questions, improving their chances of scoring high marks in the exams.

• Practising important questions helps students learn how to approach different types of problems, including derivations, conceptual questions, and application-based problems, boosting their analytical and problem-solving skills.

• The curated set of important questions eliminates the need for students to search through multiple resources. They focus on the most relevant problems, saving time and allowing targeted preparation.

• By solving these questions, students gain insights into how conic sections like parabolas, ellipses, and hyperbolas are used in real-world scenarios, such as satellite paths, bridge designs, and optics.

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CBSE Class 11 Maths Chapter-wise Important Questions

CBSE Class 11 Maths Chapter-wise Important Questions and Answers cover topics from all 14 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

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