NCERT Exemplar Solutions Class 11 Maths Conic Sections — Fill in the Blanks and Objective Type Q41–59

NCERT Exemplar Problems, Class 11 Mathematics — Conic Sections, Fill in the Blanks and Objective Type questions 41 to 59. All 19 questions solved.

Questions 41 to 46 ask for a single equation or length. Questions 47 to 59 are multiple choice: circle area and circles through a point (47 to 50), parabolas from focus and directrix (51 to 53), then the ellipse and hyperbola from eccentricity, latus rectum and foci (54 to 59). Several are shorter versions of earlier questions in this unit.

One question in this band, Question 49, is misprinted: none of its four printed options is correct, and the printed key’s choice does not satisfy the conditions. It is explained where it occurs.

Key insight. For a multiple-choice question, compute the answer from the data first and only then look at the options. Most distractors are the answer to a nearly identical question (the other axis, the other sign, $b^2$ for $a^2$). When no option matches, suspect a misprint only after you have re-checked your own working.

Fill in the blanks in Questions 41 to 46.

Question 41

The equation of the circle having centre at $(3,-4)$ and touching the line $5x+12y-12=0$ is ________.

Solution. A circle touching a line has its radius equal to the distance from the centre to that line:

$$r=\frac{|5(3)+12(-4)-12|}{\sqrt{25+144}}=\frac{|15-48-12|}{13}=\frac{45}{13}.$$

So the circle is $(x-3)^2+(y+4)^2=\left(\frac{45}{13}\right)^2$.

Question 42

The equation of the circle circumscribing the triangle whose sides are the lines $y=x+2$, $3y=4x$, $2y=3x$ is ________.

Solution. Find the vertices by pairing the lines. $y=x+2$ with $3y=4x$: $3x+6=4x$, so $(6,8)$. $y=x+2$ with $2y=3x$: $2x+4=3x$, so $(4,6)$. $3y=4x$ with $2y=3x$ meet at the origin $(0,0)$.

The circle passes through the origin, so it has the form $x^2+y^2+Dx+Ey=0$. At $(6,8)$: $100+6D+8E=0$. At $(4,6)$: $52+4D+6E=0$. Solving gives $D=-46$, $E=22$.

Question 43

An ellipse is described by using an endless string which is passed over two pins. If the axes are $6$ cm and $4$ cm, the length of the string and distance between the pins are ________.

Solution. Here $2a=6$ and $2b=4$, so $a=3$, $b=2$ and $c=\sqrt{9-4}=\sqrt5$. The pins are the foci, so the distance between them is $2c=2\sqrt5$ cm.

When the pencil pulls the loop taut at a point $P$, the string runs $S\to P\to S’\to S$, so its length is $SP+PS’+SS’=2a+2c=6+2\sqrt5$ cm.

Question 44

The equation of the ellipse having foci $(0,1)$, $(0,-1)$ and minor axis of length $1$ is ________.

Solution. The foci are on the $y$-axis, so the major axis is vertical and $c=1$. The minor axis is $1$, so $b=\frac12$ and $b^2=\frac14$. Then $a^2=b^2+c^2=\frac14+1=\frac54$ and

$$\frac{x^2}{1/4}+\frac{y^2}{5/4}=1.$$

Question 45

The equation of the parabola having focus at $(-1,-2)$ and the directrix $x-2y+3=0$ is ________.

Solution. This is Question 28(c) again. Equating the squared distance to the focus with the squared distance to the line, $(x+1)^2+(y+2)^2=\frac{(x-2y+3)^2}{5}$, and expanding:

Question 46

The equation of the hyperbola with vertices at $(0,\pm6)$ and eccentricity $\dfrac53$ is ________ and its foci are ________.

Solution. Vertices on the $y$-axis give a vertical transverse axis with $a=6$. Then $c=ae=10$ and $b^2=c^2-a^2=100-36=64$. The equation is $\frac{y^2}{36}-\frac{x^2}{64}=1$, and the foci are $(0,\pm c)$.

Choose the correct answer out of the given four options (M.C.Q.) in Questions 47 to 59.

Question 47

The area of the circle centred at $(1,2)$ and passing through $(4,6)$ is

Solution. The radius is the distance from the centre to the given point: $\sqrt{3^2+4^2}=5$. So the area is $\pi r^2=25\pi$. Options (A) and (B) are what you get by treating $r$ or $r^2=25$ carelessly (for instance $\pi\cdot5$ or $2\pi\cdot5$ for a circumference-style answer).

Question 48

Equation of a circle which passes through $(3,6)$ and touches the axes is

Solution. The point $(3,6)$ is in the first quadrant, so the circle touching both axes has centre $(r,r)$ with $r>0$. Passing through $(3,6)$:

$$(3-r)^2+(6-r)^2=r^2\ \Rightarrow\ r^2-18r+45=0\ \Rightarrow\ r=3\text{ or }15.$$

For $r=3$: $(x-3)^2+(y-3)^2=9$, i.e. $x^2+y^2-6x-6y+9=0$, which is option (C). The second circle, $r=15$, is $x^2+y^2-30x-30y+225=0$ and is not offered, so (C) is the only option that is correct.

Question 49

Equation of the circle with centre on the $y$-axis and passing through the origin and the point $(2,3)$ is

(A) $x^2+y^2+13y=0$ (B) $3x^2+3y^2+13x+3=0$ (C) $6x^2+6y^2-13x=0$ (D) $x^2+y^2+13x+3=0$

Solution. A circle through the origin with its centre on the $y$-axis has the form $x^2+y^2+Ey=0$ (no $x$ term, no constant). Passing through $(2,3)$: $4+9+3E=0$, so $E=-\frac{13}{3}$. The circle is

$$x^2+y^2-\frac{13}{3}y=0\ \Longleftrightarrow\ 3x^2+3y^2-13y=0,$$

with centre $\left(0,\frac{13}6\right)$.

None of the four printed options is this circle. Option (A) has the wrong sign of $E$ and the wrong size; (B) and (D) are not through the origin; and the printed key’s choice (C), $6x^2+6y^2-13x=0$, has its linear term in $x$, which would put the centre on the $x$-axis, and in any case fails at $(2,3)$ (it gives $24+54-26=52\ne0$). The options are misprinted in the book. The nearest in shape is (C), since the correct circle can be written $6x^2+6y^2-26y=0$, but (C) is not that circle. Since no option is right, none is marked above; the correct equation is the one in the box.

Question 50

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length $3a$ is

Solution. In an equilateral triangle the centroid is also the circumcentre, so the origin is the centroid. The centroid divides each median in the ratio $2:1$, so the circumradius is $\frac23\cdot3a=2a$ and the circle is $x^2+y^2=4a^2$. Option (A) takes the whole median as the radius.

Question 51

If the focus of a parabola is $(0,-3)$ and its directrix is $y=3$, then its equation is

Solution. The vertex is midway between focus and directrix, at the origin, and $a=3$. The focus is below the directrix, so the parabola opens downward: $x^2=-4ay=-12y$. Check with the definition: $x^2+(y+3)^2=(y-3)^2$ gives $x^2=-12y$.

Question 52

If the parabola $y^2=4ax$ passes through the point $(3,2)$, then the length of its latus rectum is

Solution. Substituting $(3,2)$: $4=4a\cdot3$, so $a=\frac13$. The latus rectum has length $4a=\frac43$. Option (A) is $2a$, the length of the half chord.

Question 53

If the vertex of the parabola is the point $(-3,0)$ and the directrix is the line $x+5=0$, then its equation is

Solution. The distance from vertex $(-3,0)$ to the directrix $x=-5$ is $a=2$. The directrix is to the left of the vertex, so the parabola opens to the right: $y^2=4a(x+3)=8(x+3)$. The directrix is vertical, so the equation has $y^2$, which rules out (B).

Question 54

The equation of the ellipse whose focus is $(1,-1)$, the directrix the line $x-y-3=0$ and eccentricity $\dfrac12$ is

Solution. Distance to the focus is $e$ times distance to the directrix. Squaring,

$$(x-1)^2+(y+1)^2=\frac14\cdot\frac{(x-y-3)^2}{2}\ \Rightarrow\ 8\left[(x-1)^2+(y+1)^2\right]=(x-y-3)^2.$$

Expanding: $8x^2+8y^2-16x+16y+16=x^2+y^2+9-2xy-6x+6y$, so $7x^2+2xy+7y^2-10x+10y+7=0$. The options differ only in the linear and constant terms, so the signs of $-10x+10y$ decide it.

Question 55

The length of the latus rectum of the ellipse $3x^2+y^2=12$ is

Solution. Divide by $12$: $\frac{x^2}{4}+\frac{y^2}{12}=1$. The larger denominator is under $y^2$, so $a^2=12$ (major axis vertical) and $b^2=4$. The latus rectum is $\frac{2b^2}{a}=\frac{8}{2\sqrt3}=\frac{4}{\sqrt3}$. Swapping the denominators (taking $a^2=4$, $b^2=12$) gives the wrong value $\frac{2\cdot12}{2}=12$, which is not among the options either.

Question 56

If $e$ is the eccentricity of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a<b)$, then

Solution. The condition $a<b$ means the major axis is along the $y$-axis, so the larger quantity is $b$, and the usual formula “(smaller)$^2$ = (larger)$^2(1-e^2)$” reads $a^2=b^2(1-e^2)$. Options (C) and (D) have $e^2-1<0$, which would make a square negative.

Question 57

The eccentricity of the hyperbola whose latus rectum is $8$ and conjugate axis is equal to half of the distance between the foci is

Solution. The conjugate axis is $2b$ and the distance between the foci is $2c$, so $2b=\frac12(2c)=c$, that is $c=2b$. Then $c^2=a^2+b^2$ gives $4b^2=a^2+b^2$, so $a^2=3b^2$ and

$$e=\frac ca=\frac{2b}{\sqrt3\,b}=\frac{2}{\sqrt3}.$$

The latus rectum $8$ is not needed for $e$; it fixes the size ($a=12$, $b^2=48$) but not the shape, and the eccentricity depends only on shape.

Question 58

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt2$. Its equation is

Solution. This is Question 20: $c=8$, $a^2=32$, $b^2=32$, so $x^2-y^2=32$. Option (B) has $e^2=1+\frac94=\frac{13}{4}$, and (C) is not even a conic with two squared terms.

Question 59

Equation of the hyperbola with eccentricity $\dfrac32$ and foci at $(\pm2,0)$ is

Solution. This is Question 22: $c=2$, $a=\frac43$, $a^2=\frac{16}9$ and $b^2=\frac{20}9$, so $\frac{9x^2}{16}-\frac{9y^2}{20}=1$. Multiplying by $\frac49$ gives $\frac{x^2}{4}-\frac{y^2}{5}=\frac49$, option (A). Option (C) has $c^2=13$, so its foci are not at $(\pm2,0)$.

Common mistakes

• Q42: Writing the circle as $x^2+y^2+Dx+Ey+F=0$ and forgetting that one vertex is the origin, so $F=0$; with three unknowns instead of two the algebra becomes much longer.
• Q43: Giving the string length as $2a=6$. A loop passed over the pins also runs along the pin-to-pin segment, so it is $2a+2c$.
• Q44: Reading “minor axis of length $1$” as $b=1$. The length is $2b$, so $b=\frac12$.
• Q48: Using the centre $(-r,-r)$ or $(r,-r)$ for a point in the first quadrant. The sign of the centre follows the quadrant of the point.
• Q55, Q56: Taking $a^2$ as the denominator under $x^2$ every time. In $3x^2+y^2=12$ and in $a<b$ the major axis lies on the $y$-axis, and the larger denominator is the one that plays the role of $a^2$ in $b^2=a^2(1-e^2)$.
• Q57, Q59: Using the ellipse relation $b^2=a^2-c^2$ for a hyperbola. A hyperbola has $c^2=a^2+b^2$, so $c$ is the largest of the three.

Practise next

• Other pages in this unit: Short Answer Type, Q1–22 and Long Answer Type and True or False, Q23–40.
• NCERT textbook: Exercise 10.1 for circles, Exercise 10.2 for parabolas and the Miscellaneous Exercise.