Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
The center of the ellipse $\dfrac{(x+y-2^2)}{9}+\dfrac{(x-y)^2}{16}=1 $ is________
- $ (0,0) $
- $(0, 1 )$
- $ (1, 0) $
- $ (1 , 1 ) $
Correct

Incorrect

Hint

The center of the given ellipse is the point of intersection of the lines $x+y-2 =0$ and $x-y =0$ . After solving above two equations we get $x=1$ and $y=1$ .
$\therefore$ Centere $= (1 ,1 )$
Question 2 of 25

2. Question
1 point(s)
The parametric co-ordinate of any of the parabola $ y^2 =4ax $ is_______
- $ (-at^2 , -2at) $
- $(at^2, 2at) $
- $ (a \sin^2 t, -2a\sin t) $
- $ (a \sin^2 t, 2a\sin t) $
Correct

Incorrect

Hint

$y^2=4ax$ ( Equation of Parabola)
The parametric form of this parabola is $x=at^2 , y=2at$
$\therefore$ Coordinate of any points is $(at^2, 2at )$
Question 3 of 25

3. Question
1 point(s)
The equation of parabola with vertex at origin the axis is along $ x $-axis and passing through the point $ (2, 3 ) $ is_____
- $ y^2 =9x $
- $y^2 = \dfrac{9x}{2} $
- $ y_2=2x $
- $ y^2=dfrac{9x}{4} $
Correct

Incorrect

Hint

Equation of the parabola is $y^2= 4ax$ (i)
Given point $= (2, 3 )$
$\implies 3^2 =4ax^2$
$\implies 4a= \dfrac{9}{2}$
Putting the value of $4a$ in (i).
$\therefore y^2=\dfrac{9}{2} x$
Question 4 of 25

4. Question
1 point(s)
At what point of the parabola $x^2=ay $ is the abscissa three times that of ordinate ______
- $(1, 1) $
- $ (3, 1) $
- $ (-3,1) $
- $(-3,-3 ) $
Correct

Incorrect

Hint

Given, Parabola is $x^2 =ay$
Let $P (h, k)$ be the point on the parabola.
So $h=3k$ ( According to question )
Since $P (h,k)$ lies on the parabola $\implies h^2=9k$ .
From above two equations we get $(3k)^2 =9k$
$\implies 9k^2=9k \implies 9k^2-9k =0$
$\implies 9 k (k-1) =0 \implies k=0 ,1$
When $k=0, h=0$
When $k=1, h=3$
$\therefore P (h, k ) =(3,1 )$
Question 5 of 25

5. Question
1 point(s)
In an ellipse the distance between its foci is $ 6 $ and its minor axis is $ 8 $ then $ e $ is ________
- $ \dfrac{4}{5} $
- $\dfrac{1}{\sqrt{52}} $
- $ \dfrac{3}{5} $
- $\dfrac{1}{2} $
Correct

Incorrect

Hint

Given, distance between foci $=6$
$\implies 2ae=6$
$\implies ae=3$
Minor axis $=8$
$\implies 2b=8 \implies b=4$
$\implies b^2=16 \implies a^2(1-e^2)=16$
$\implies a^2-a^2e^2=16 \implies a^2-(ae)^2=16$
$\implies a^2-3^2=16 \implies a^2-9=16$
$\implies a^2=25 \implies a=5$
Now, $ae=3 \implies 5e=3$
$\implies e= \dfrac{3}{5}$
Question 6 of 25

6. Question
1 point(s)
The equation of parabola whose focus is $ (3 , 0) $ and directrix is $ 3x+4y =1 $______
- $16x^2-9y^2-24xy-144x+8y+224=0 $
- $16x^2+9y^2-24xy-144x+8y-224=0 $
- $16x^2+9y^2-24xy-144x+8y+224=0 $
- $ 16x^2+9y^2+24xy+144x+8y+224=0 $
Correct

Incorrect

Hint

Given, $F= (3 , 0)$ , Equation of directrix $: 3x+4y=1$
$\implies 3x+4y-1=0$
Let $P (x, y)$ be any point on parabola and $PM$ be the length of perpendicular.
$\implies FP= PM \implies FP^2 =PM^2$
$\implies (x-3)^2 +(y-0)^2= \left\{ \dfrac{(3x+4y-1)}{\sqrt{3^2+4^2}} \right\}^2$
$\implies x^2+9-6x+y^2=\dfrac{9x^2+16y^2+1+24xy-8y-6x}{25}$
$\implies 25x^2+225-150x+25y^2=9x^2+16y^2+24xy-8y-6x+1$
$\implies 16x^2+9y^2-24xy -144x+8y+224=0$
Question 7 of 25

7. Question
1 point(s)
The parametric representation of $ ( 2+t^2 , 2t+1) $ represents_____
- a parabola
- a hyperbola
- an ellipse
- a circle
Correct

Incorrect

Hint

$x =2+t^2 \implies t^2 =x-2$ (i)
$y=2t+1 \implies 2t=y-1\implies t= \dfrac{y-1}{2}$ (ii)
From (i) and (ii)
$x-2 = \left( \dfrac{y-1}{2} \right)^2$
$\implies x-2= \dfrac{(y-1)^2}{4}$
$\implies (y-1)^2= 4(x-2)$
This represents the equation of parabola.
Question 8 of 25

8. Question
1 point(s)
The equation of a hyperbola with foci on the $ x $-axis is_____
- $ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 $
- $ \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 $
- $x^2+y^2=(a^2+b^2) $
- $x^2-y^2=(a^2+b^2) $
Correct

Incorrect

Hint

Equation of hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
Question 9 of 25

9. Question
1 point(s)
The line $ lx+my+n=0 $ will touch the parabola $ y^2=4ax $ if_______
- $ ln=am^2 $
- $ln=am $
- $ ln=a^2m^2 $
- $ln=a^2m $
Correct

Incorrect

Hint

Given, $lx+my+n=0$
$\implies my=-lx-n$
$\implies y= \dfrac{-lx-n}{m}$
$\implies y= \left( -\dfrac{l}{m} \right)x+\left(- \dfrac{n}{m} \right)$
This will touch the parabola $y^2=4ax$ if $\left( -\dfrac{n}{m} \right)= \dfrac{a}{\left( -\frac{l}{m} \right)}= -\dfrac{am}{l}$
$\implies ln =am^2$
Question 10 of 25

10. Question
1 point(s)
A rod of length $ 12\operatorname {cm} $ moves with its and always touching co-ordinate axes. Then the equation of the locus of a point $ P $ on the rod which $ 3 \operatorname {cm} $ from the end in contact with the $ x $-axis is_______
- $\dfrac{x^2}{81}+\dfrac{y^2}{4}=1 $
- $\dfrac{x^2}{81}+\dfrac{y^2}{9}=1 $
- $ \dfrac{x^2}{169}+\dfrac{y^2}{9}=1 $
- $\dfrac{x^2}{9}+\dfrac{y^2}{81}=1 $
Correct

Incorrect

Hint

Hence $AP = 3 \operatoname {cm}, AB =12 \operatorname {cm}$
Now $BP =AB-AP = 12-3 =9 \operatorname {cm}$
Since $PQ || OA \implies m < PAO =m < BPO$
Now in $\triangle BPO, \cos \theta = \dfrac{QP}{BP}$
$\implies \cos \theta = \dfrac{x}{9}$
Again in $\triangle PAR, \sin \theta = \dfrac{PR}{PA}=\dfrac{y}{3}$
From above two equations , we get $\sin ^2 \theta +\cos^2 \theta =\left( \dfrac{y}{3} \right)^2+\left( \dfrac{x}{9} \right)^2$
$\implies \dfrac{x^2}{81}+\dfrac{y^2}{9}=1\quad (\text { Equation of Ellipse})$
Question 11 of 25

11. Question
1 point(s)
If $ (a, b) $ is the midpoint of a chord passing through the vertex of the parabola $ y^2=4x $, then______
- $ a=2b $
- $ 2a=b $
- $a^2=2b $
- $ 2a=b^2 $
Correct

Incorrect

Hint

Let $P (x, y)$ be the other end of the chord $OP$ of $O (0,0)$ .
Now, $\dfrac{(x+0)}{2}=a$
$\implies x=2a$
And $\dfrac{(y+0)}{2} =b$
$\implies y=2b$
$y^2=4x$
$\implies (2b)^2 =4 \times 2a$
$\implies 4b^2=8a$
$\implies b^2=2a$
Question 12 of 25

12. Question
1 point(s)
The equation of parabola with vertex $ (-2, 1 ) $ and focus $(-2 ,4) $ is______
- $ 10y= x^2+4x+16 $
- $12y= x^2+4x+16 $
- $ 12y= x^2+4x $
- $12y= x^2+4x+8 $
Correct

Incorrect

Hint

Given, vertex $= (-2 ,1)$ , Focus $= (-2 , 4)$
Parabola axis of symmetry as $x=-2$
$\implies x+2=0$
Hence, the equation of the parabola is of the type $y-k=a (x-h)^2$ where $(h , k)$ is vertex.
Now focus $= (h ,k + \dfrac{1}{4a})$
So, focus $=(-2, 1 +\dfrac{1}{4a})$
$\implies 1 + \dfrac{1}{4a}=4$
$\implies \dfrac{1}{4a}=3 \implies 4a =\dfrac{1}{3}\implies a =\dfrac{1}{12}$
Now equation of parabola is $(y-1)= \left( \dfrac{1}{12} \right)(x+2)^2$
$\implies 12(y-1)= (x+2)^2$
$\implies 12y=x^2+4x+16$
Question 13 of 25

13. Question
1 point(s)
If a parabolic reflector is $ 20 \operatorname {cm} $ in diameter and $5 \operatorname {cm} $ deep then the focus of parabolic reflector is_______
- $(0, 0) $
- $ ( 0, 5) $
- $(5, 0) $
- $ (5 , 5) $
Correct

Incorrect

Hint

Given, diameter of the parabola is $20 \operatorname { m}$ .
Equation of parabola is $y^2=4ax$
Parabola passes through the point $( 5 , 10)$
$10^2 =4a \times 5 \implies 100 =20a$
$\implies a = \dfrac{100}{200}=5$
So, focus $= (a, 0) = (5, 0)$
Question 14 of 25

14. Question
1 point(s)
The parametric form of the ellipse $4(x+1)^2+ (y-1)^2=4 $ is_______
- $x =\cos \theta -1, \quad y=2\sin \theta -1 $
- $ x =2\cos \theta -1, \quad y=\sin \theta +1 $
- $ x =\cos \theta -1, \quad y=2\sin \theta +1 $
- $ x =\cos \theta +1, \quad y=2\sin \theta +1 $
Correct

Incorrect

Hint

Given, $4(x+1)^2+ (y-1)^2=4$
$\implies (x+1)^2+ \dfrac{(y-1)^2}{4}=1$
$\therefore$ Parametric equation of given ellipse is $x+1=\cos \theta$ and $y-1=2 \sin \theta$
$\implies x =\cos \theta -1, \quad y=2 \sin \theta +1$
Question 15 of 25

15. Question
1 point(s)
A man running a race course notes that the sum of the distance from the two flag posts from him is always $ 10 \operatorname {meter} $ and the distance between the flag posts is $8 \operatorname { meter} $. The equation of the posts traced by the man is_________
- $\dfrac{x^2}{9}+ \dfrac{y^2}{5}=1 $
- $ \dfrac{x^2}{25}+ \dfrac{y^2}{9}=1 $
- $ \dfrac{x^2}{25}- \dfrac{y^2}{9}=1 $
- $\dfrac{x^2}{9}+ \dfrac{y^2}{16}=1 $
Correct

Incorrect

Hint

Let the equation of ellipse is $\dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2}=1$
Given, $PF+PF'=10$
$\implies 2a=10 \implies a=5$
Given, $2ae=8 \implies ae=4$
Now, $b^2=a^2 (1-e^2)$
$\implies b^2=a^2-a^2e^2 \implies b^2=a^2-(ae)^2$
$\implies b^2=5^2-4^2\implies b^2=25-16=9$
$\implies b=3$
$\therefore$ Equation of the ellipse is $\dfrac{x^2}{5^2}+ \dfrac{y^2}{3^2}=1 \implies \dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Question 16 of 25

16. Question
1 point(s)
Let $ O $ be the vertex and $ Q $ be any point on the parabola $x^2=8y $. If the point $ P $ divides the line segment $OQ $ internalyin ratio $ 1:3 $ , then the locus of $ P $ is_______
- $ x^2=y $
- $ y^2=x $
- $y^2=2x $
- $ x^2=2y $
Correct

Incorrect

Hint

Given, $x^2=8y$
Let $Q= (4t , 2t^2)$
Let $P (h , k)$ divides $\overline{OQ}$ joining $(0 ,0 )$ and $(4t ,2t^2)$ in ratio $1 : 3$ .
$\therefore h=\dfrac{1 \times 4t+3 \times 0}{4}\implies4h =4t$
$\implies h=t$
and $k=\dfrac{1\times 2t^2+3 \times0}{4}\implies4k=2t^2$
$\implies k=\dfrac{t^2}{2}$
$\implies k=\dfrac{h^2}{2}\implies2k=h^2$
$\implies x^2=2y$ , required locus.
Question 17 of 25

17. Question
1 point(s)
The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $ y^2=4ax $ is another parabola with directrix, is_______________
- $x=-a $
- $ x=-\dfrac{a}{2} $
- $x=0 $
- $ x=\dfrac{a}{2} $
Correct

Incorrect

Hint

Let $P (h, k)$ be midpoint of the line segment joining the focus $(a, 0)$ and $Q (x, y)$ . Then, $h = \dfrac{x+a}{2}, \quad k=\dfrac{y}{2}$
$\implies x=2h-a, \quad y=2k$
Given, $y^2=4ax$
$\implies (2k)^2=4(2h-a)a$
$\implies 4k^2=8ha-4a^2$
$\implies k^2=2ah-a^2$
Locus of $P(h, k)$ is $y^2=ax-a^2$
$\implies y^2=2a \left( x-\dfrac{a}{2} \right)$
Its directrix is $x-\dfrac{a}{2}=-\dfrac{a}{2}$
$\implies x=0$
Question 18 of 25

18. Question
1 point(s)
The equation of the directrix of the parabola $ y^2+4y+4x+2=0 $ is_______
- $ x=-1 $
- $x=1 $
- $x=-\dfrac{3}{2} $
- $x= \dfrac{3}{2} $
Correct

Incorrect

Hint

Given, $y^2+4y+4x+2=0$
$\implies y^2+4y+4+4x-2=0$
$\implies (y+2)^2=-4x+2$
$\implies (y+2)^2=-4\left( x-\dfrac{1}{2} \right)$
Replacing $y+2=y, \quad x-\dfrac{1}{2}=x$
We have, $y^2=4x$
This is the parabola with directrix at $x=1$
$\implies x-\dfrac{1}{2}=1$
$\implies x=1+\dfrac{1}{2}$
$\implies x= \dfrac{3}{2}$
Question 19 of 25

19. Question
1 point(s)
If the line $x-1=0 $ is the directrix of the parabola $y^2-kx+8=0 $, then value of $ k $ is_______
- $ \dfrac{1}{8} $
- $8 $
- $4 $
- $\dfrac{1}{4} $
Correct

Incorrect

Hint

Given, $y^2=kx-8$
$\implies = k\left( x-\dfrac{8}{k} \right)$
Shifting the origin, where $y=y, \quad x=x-\dfrac{8}{k}$
Directrix of standard parabola is $x=-\dfrac {k}{4}$
Directrix of original parabola is $x= \dfrac{8}{k}-\dfrac{k}{4}$
Now, $x=1$ also coincides with $x=\dfrac{8}{k}-\dfrac{k}{4}$ on solving this we get $k=4$ .
Question 20 of 25

20. Question
1 point(s)
The radius of a circle having minimum area, which touches the curve $ y=4-x^2 $ and the lines $ y=|x| $ is_______
- $ 2 (\sqrt{2}+1) $
- $2 (\sqrt{2}-1) $
- $4 (\sqrt{2}-1) $
- $4 (\sqrt{2}+1) $
Correct

Incorrect

Hint

Let radius of circle is $r$ .
Co-ordinate of centre is $(0, 4-r)$
Since, circle touches the line $y=x$ in $1{\text {st}}$ quadrant.
$\therefore \left| \dfrac{0-(4-r)}{\sqrt{2}} \right|=r \implies r-4=\pm r\sqrt{2}$
$\implies r=\dfrac{4}{\sqrt{2}+1} \quad\text {or}\quad \dfrac{4}{-\sqrt{2}+1}$
But $r \neq \dfrac{4}{1-\sqrt{2}}\left( \therefore\dfrac{4}{1-\sqrt{2}} < 0\right)$
$\therefore r= \dfrac{4}{\sqrt{2}+1}=4(\sqrt{2}-1)$
Question 21 of 25

21. Question
1 point(s)
The slope of the line touching both the parabolas $ y^2=4x $ and $x^2=-32y $ is________
- $ \dfrac{1}{2} $
- $\dfrac{3}{2} $
- $\dfrac{1}{8} $
- $ \dfrac{2}{3} $
Correct

Incorrect

Hint

Let the tangent to the parabola be $y=mx+\dfrac{a}{m}$ , if it touches the curve, then $D=0$ , to get the value of $m$ .
For parabola , $y^2=4x$ .
Let $y=mx+ \dfrac{1}{m}$ be tangent line and it touches the parabola $x^2=-32y$
$\implies x^2=-32 \left( mx+\dfrac{1}{m} \right)$
$\implies x^2+32mx +\dfrac{32}{m}=0$
$\implies D=0$
$\therefore (32m)^2-4 \left( \dfrac{32}{m} \right)=0 \implies m^3= \dfrac{1}{8}$
$\therefore m=\dfrac{1}{2}$
Question 22 of 25

22. Question
1 point(s)
The tangent at $ (1 ,7) $ to the curves $ x^2=y-6x $ touches the circle $x^2+y^2+16x+12y+c=0 $ at_______
- $(6 , 7) $
- $ (-6, 7) $
- $(6,-7) $
- $(-6,-7) $
Correct

Incorrect

Hint

The tangent at $(1, 7)$ to the parabola $x^2=y-6x$ is $x (1) =\dfrac{1}{2}(y+7)-6$ [ replacing mathjax] $x\to xx_1$ and $2y \to y+y_1$ ]
$\implies 2x=y+7-12$
$\implies y=2x+5$
Which is also tangent to the circle $x^2+y^2+16x+12y+c=0$
i.e., $x^2+(2x+5)^2+16x+12 (2x+5)+c=0$
It must have two equal roots ( $\alpha =\beta$ )
$\implies 5x^2+60x+85+c=0$
$\implies \alpha+\beta = -\dfrac{60}{5} \implies \alpha=-6$
$\implies x=-6$ and $y=2x+5 =-7$
$\therefore$ Point of contact is $(-6, -7)$
Question 23 of 25

23. Question
1 point(s)
The angle between the tangents drawn from the point $(1 , 4) $ to the parabola $ y^2=4x $ is________
- $ \dfrac{\pi}{6} $
- $ \dfrac{\pi}{4} $
- $ \dfrac{\pi}{3} $
- $ \dfrac{\pi}{2} $
Correct

Incorrect

Hint

Tangent to $y^2=4ax$ is $y=mx+\dfrac{a}{m}$
$\implies$ Given $y^2=4x$ , tangent $: y=mx +\dfrac{1}{m}$
Since, tangent passes through $(1, 4)$
$\implies 4=m+\dfrac{1}{m} \implies m^2-4m+1=0$
$\therefore m_1+m_2=4$ and $m_1m_2=1$
$\implies |m_1-m_2| =\sqrt{4^2-4\cdot1}=\sqrt{12}=2\sqrt{3}$
$\therefore \tan \theta =\left| \dfrac{m_1-m_2}{1+m_1m_2} \right|=\left| \dfrac{2\sqrt{3}}{1+1} \right|=\sqrt{3}$
$\implies \theta = \tan ^{-1} \sqrt{3}=\dfrac{\pi}{3}$
Question 24 of 25

24. Question
1 point(s)
The focal chord to $ y^2=16x $ is tangent to $ (x-6)^2+y^2=2 $, then the possible values of the slope of this chord are_______
- $\{-1,1\} $
- $\{-2,2\} $
- $\left\{ -2, \dfrac{1}{2} \right\} $
- $\left\{ 2, -\dfrac{1}{2} \right\} $
Correct

Incorrect

Hint

Given, $y^2=16x$ and $(x-6)^2+y^2=2$
$\implies F =(4 , 0)$
Now, tangents are drawn from $(4, 0)$ to $(x-6)^2+y^2=2$
Since, $FA$ is tangent to circle.
$\therefore \tan \theta= \text { Slope of tangent} =\dfrac{AC}{AF}=\dfrac{\sqrt{2}}{\sqrt{2}}=1 \quad \text {or} \quad\dfrac{BC}{BF}=-1$
$\therefore$ Slope of the focal chord as tangent to circle $= \{-1, 1\}$
Question 25 of 25

25. Question
1 point(s)
The equation of the common tangent to the curves $ y^2=8x $ and $ xy=-1 $ is_______
- $ 3y=9x+2 $
- $y=2x+1 $
- $2y=x+8 $
- $ y=x+2 $
Correct

Incorrect

Hint

Tangent to the curve $y^2=8x$ is $y=mx+\dfrac{2}{m}$ . So it must satisfy $xy=-1$ .
$\implies x\left( mx+\dfrac{2}{m} \right)=-1 \implies mx^2+\dfrac{2}{m}x+1=0$
Since $D=0$
$\implies \dfrac{4}{m^2}=4m=0$
$\implies m^2=1$
$\implies m=1$
Equation of common tanget is $y=x+2$

Mathematics Class XI

Practice Questions 1-Conic Sections-Class XI

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