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Maths M.C.Q (1 Marks)

Ellipse · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 10 questions.

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M.C.Q (1 Marks)

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10 questions · auto-graded multiple-choice test.

MCQ 11 Mark
The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus-rectum, is:
  • ✓
    $\frac{\sqrt{5}-1}{2}$
  • B
    $\frac{\sqrt{5}+1}{2}$
  • C
    $\frac{\sqrt{5}-1}{4}$
  • D
    $\text{none of these}$
Answer
Correct option: A.
$\frac{\sqrt{5}-1}{2}$
According to the question, the distance between the foci is equal to the length of the latus rectum.
$\frac{2\text{b}^2}{\text{a}}=2\text{ae}$
$\Rightarrow\text{b}^2=\text{a}^2\text{e}$
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{\text{a}^2\text{e}}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\text{e}}$
On squaring both sides, we get:
$\text{e}^2\text{e}-1=0$
$\Rightarrow\text{e}=\frac{-1\pm\sqrt{1+4}}{2}$
$\Rightarrow\text{e}=\frac{\sqrt{5}-1}{2}$ $(\because\text{e}\text{ cannot be negative}\big)$
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MCQ 21 Mark
The eccentricity of the conic $9\text{x}^2+25\text{y}^2=225$ is:
  • A
    $\frac{2}{5}$
  • ✓
    $\frac{4}{5}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{5}$
Answer
Correct option: B.
$\frac{4}{5}$
$\Rightarrow\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
Comparing it with $\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ we get:
$\text{a}=5$ and $\text{b}=3$
Here, a > b, so the major and the minor axes of the ellipse are along the x−axis and y−axis, respectively.
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{9}{25}}$
$\Rightarrow\text{e}=\sqrt{\frac{16}{25}}$
$\Rightarrow\text{e}=\frac{4}{5}$
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MCQ 31 Mark
The difference between the lengths of the major axis and the latus$-$rectum of an ellipse is
  • A
    $ae$
  • B
    $2ae$
  • C
    $ae^2$
  • ✓
    $2ae^2$
Answer
Correct option: D.
$2ae^2$
Length of the latus rectum $=\frac{2\text{b}^2}{\text{a}}$
and $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\text{a}^2\text{e}^2=\text{a}^2-\text{b}^2$
$\Rightarrow\text{b}^2=\text{a}^2-\text{a}^2\text{e}^2$
$\Rightarrow\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\therefore\ $Length of the latus rectum $=\frac{2\text{a}^2(1-\text{e}^2)}{\text{a}}=2\text{a}(1-\text{e}^2)$
Length of the major axis $= 2a$
Difference between length of latus rectum and length of major axis $=2\text{a}-2\text{a}(1-\text{e}^2)$
$\\=2\text{a}-2\text{a}+2\text{ae}^2\\=2\text{ae}^2$
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MCQ 41 Mark
The equations of the tangents to the ellipse $9\text{x}^2+16\text{y}^2=144$ from the point (2, 3) are:
  • A
    y = 3, x = 5
  • B
    x = 2, y = 3
  • C
    x = 3, y = 2
  • ✓
    x + y = 5, y = 3
Answer
Correct option: D.
x + y = 5, y = 3
$\Rightarrow\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1$
Equation of the tangent in case of an ellipse is given by
$\text{y}=\text{mx}+\sqrt{\text{a}^2\text{m}^2+\text{b}^2}$
$\Rightarrow\text{y}=\text{mx}+\sqrt{16\text{m}^2+9}\ \dots(1)$
Substituting x = 2 and y = 3, we get:
$3=2\text{m}\pm\sqrt{16\text{m}^2+9}$
$\Rightarrow3-2\text{m}=\sqrt{16\text{m}^2+9}$
On squaring both sides, we get:
$(3-2\text{m})^2=(16\text{m}^2+9)$
$\Rightarrow9+4\text{m}^2-12\text{m}=(16\text{m}^2+9)$
$\Rightarrow12\text{m}^2+12\text{m}=0$
$\Rightarrow12\text{m}(\text{m+1})=0$
$\Rightarrow\text{m}=0,-1$
Substituting values of m in eq. (1), we get:
For $\text{m}=0,\ \text{y}=3$
For $\text{m}=-1,\ \text{y}=-\text{x}+5$ or $\text{x}+\text{y}=5$
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MCQ 51 Mark
The eccentricity of the ellipse $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{y}^2}=1$ if its latus rectum is equal to one half of its minor axis, is:
  • A
    $\frac{1}{\sqrt{2}}$
  • ✓
    $\frac{\sqrt{3}}{2}$
  • C
    $\frac{1}{2}$
  • D
    $\text{none of these}$
Answer
Correct option: B.
$\frac{\sqrt{3}}{2}$
According to the question, the latus rectum is half its minor axis.
i.e. $\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$
$\Rightarrow2\text{b}^2=\text{ab}$
$\Rightarrow\text{a}=2\text{b}$
Now, $\text{e}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{\text{b}^2}{4\text{b}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{1}{4}}$
$\Rightarrow\text{e}=\frac{\sqrt{3}}{2}$
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MCQ 61 Mark
For the ellipse $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0$
  • A
    centre is (-1, 2)
  • B
    lengths of the axes are $\sqrt{3}$ and 1
  • C
    eccentricity $=\sqrt{\frac{2}{3}}$
  • ✓
    all of these.
Answer
Correct option: D.
all of these.
Disclaimer : The equation should be $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$ instead of $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0.$
$12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$
$\Rightarrow12\big(\text{x}^2+2\text{x}\big)+4\big(\text{y}^2-4\text{y}\big)=-24$
$\Rightarrow12\big(\text{x}^2+2\text{x}+1\big)+4\big(\text{y}^2-4\text{y}+4\big)=-24+12+16$
$\Rightarrow12\big(\text{x}+1\big)^2+4\big(\text{y}-2\big)^2=4$
$\Rightarrow\frac{(\text{x}+1)^2}{3}+\frac{(\text{y}-2)^2}{1}=1$
So, the centre is a $(-1,\ 2).$
Here, $\text{a}=\sqrt{3}$ and $\text{b}=1$
The lengths of the axes are $\sqrt{3}$ and 1.
Now, $\text{e}=\sqrt{1-\frac{\text{b}62}{\text{a}^2}}$
$\text{e}=\sqrt{1-\frac{1}{3}}$
$\Rightarrow\text{e}=\sqrt{\frac{2}{3}}$
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MCQ 71 Mark
The equation of the circle drawn with the two foci of $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ end-point of a diameter is
  • A
    $\text{x}^2+\text{y}^2=\text{a}^2+\text{b}^2$
  • B
    $\text{x}^2+\text{y}^2=\text{a}^2$
  • C
    $\text{x}^2+\text{y}^2=2\text{a}^2$
  • ✓
    $\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$
Answer
Correct option: D.
$\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$
We have $\text{r}=\text{ae}$
Let the equation of the circle be $\text{x}^2+\text{y}^2=\text{r}^2.$
Now, $\text{x}^2+\text{y}^2=\text{a}^2\text{e}^2$ $(\because\text{r}=\text{ae})$
$\Rightarrow\text{x}^2+\text{y}^2=\text{a}^2\Big(1-\frac{\text{b}^2}{\text{a}^2}\Big)$ $\bigg(\because\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}\bigg)$
$\Rightarrow\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$
$\therefore\ $The required equation of the circle $\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2.$
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MCQ 81 Mark
The eccentricity of the ellipse, if the minor axis is equal to the distance between the foci, is:
  • A
    $\frac{\sqrt{3}}{2}$
  • B
    $\frac{2}{\sqrt{3}}$
  • ✓
    $\frac{1}{\sqrt{2}}$
  • D
    $\frac{\sqrt{2}}{3}$
Answer
Correct option: C.
$\frac{1}{\sqrt{2}}$
According to the question, the minor axis is equal to the distance between the foci.
i.e. $2\text{b}=2\text{ae}$
$\text{e}=\frac{\text{b}}{\text{a}}\ \dots(1)$
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\text{e}^2}$ $\Big[\text{From}\ (1)\Big]$
On squaring both sides, we get:
$\text{e}^2=1-\text{e}^2$
$\Rightarrow2\text{e}^2=1$
$\Rightarrow\text{e}^2=\frac{1}{2}$
$\Rightarrow\text{e}=\frac{1}{\sqrt{2}}$
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MCQ 91 Mark
The latus-rectum of the conic $3\text{x}^2+4\text{y}^2-6\text{x}+8\text{y}-5=0$ is:
  • ✓
    $3$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\frac{2}{\sqrt{3}}$
  • D
    $\text{none of these}$
Answer
Correct option: A.
$3$
$\Rightarrow3(\text{x}^2-2\text{x})+4(\text{y}^2+2\text{y})=5$
$\Rightarrow3(\text{x}^2-2\text{x}+1)+4(\text{y}^2+2\text{y}+1)=5+3+4$
$\Rightarrow3(\text{x}-1)^2+4(\text{y}+1)^2=12$
$\frac{(\text{x-1})^2}{4}+\frac{(\text{y}+1)^2}{3}=1$
So, $\text{a}=2$ and $\text{b}=\sqrt{3}$
$\therefore\ $Latus rectum $=\frac{2\text{b}^2}{\text{a}}$
$\\=2\frac{\big[\sqrt{3}\big]^2}{2}\\=3$
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MCQ 101 Mark
The equation of the ellipse with focus (-1, 1), directrix x - y + 3 = 0 and eccentricity $\frac{1}{2}$ is:
  • A
    $7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}+10\text{y}+7=0$
  • ✓
    $7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}-10\text{y}+7=0$
  • C
    $7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}-10\text{y}-7=0$
  • D
    $\text{None of these}$
Answer
Correct option: B.
$7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}-10\text{y}+7=0$


Let P(x,y) be any point on the ellipse whose focus and eccentricity are S(-1,1) and $\text{e}=\frac{1}{2},$respectively.Let PM be the perpendicular from P on the directrix.
Then $\text{SP}=\text{e}\times\text{PM}$
$\Rightarrow\text{SP}=\frac{1}{2}\times\text{PM}$
$\Rightarrow2\text{SP}=\text{PM}$
$\Rightarrow4(\text{SP})^2=\text{PM}^2$
$\Rightarrow4\Big[(\text{x}+1)^2+(\text{y}-1)^2\Big]=\bigg(\frac{\text{x}-\text{y}+3}{\sqrt{1^2+}(-1)^2}\bigg)^2$
$\Rightarrow4\big[\text{x}^2+1+2\text{x}+\text{y}^2+1-2\text{y}\big]\\=\frac{{\text{x}^2+\text{y}^2+9-2\text{xy}-6\text{y}+6\text{x}}}{2}$
$\Rightarrow8\text{x}^2+8+16\text{x}+8\text{y}^2+8-16\text{y}\\=\text{x}^2+\text{y}62+9-2\text{xy}-6\text{y}+6\text{x}$
$\therefore7\text{x}^2+7\text{y}^2+2\text{xy}-10\text{y}+10\text{x}+7=0$
This is the required equation of the ellipse.
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