NCERT Exemplar Solutions Class 11 Maths Conic Sections — Long Answer Type and True or False Q23–40

NCERT Exemplar Problems, Class 11 Mathematics — Conic Sections, Long Answer Type and True or False questions 23 to 40. All 18 questions solved.

Questions 23 to 27 ask for circles from indirect data: a centre found as an intersection of lines or on a given line, a radius found from a chord or a tangency. Questions 28 to 32 ask for parabola, ellipse and hyperbola equations from the definition (distance to focus against distance to directrix, sum or difference of focal distances). Questions 33 to 40 are True or False, each to be justified.

Two points in the printed key need care: the key’s answer to Question 32(a) is wrong, and its answer to Question 27 gives only one of the two possible circles. Both are explained where they occur.

Key insight. Almost every locus question is “the set of points whose distances satisfy a rule” and the rule is the definition of the curve: sum of distances to two points is constant for an ellipse, difference for a hyperbola, and distance to a point equals $e$ times distance to a line for all three. Write the rule, square carefully, simplify, and then read $a$, $b$, $c$ off the result.

Question 23

If the lines $2x-3y=5$ and $3x-4y=7$ are the diameters of a circle of area $154$ square units, then obtain the equation of the circle.

Solution. Two diameters meet at the centre, so solve the pair: from $2x-3y=5$, $x=\frac{5+3y}{2}$; substituting in $3x-4y=7$ gives $\frac{15+9y}{2}-4y=7$, so $15+y=14$, $y=-1$ and $x=1$. Centre $(1,-1)$.

The area is $\pi r^2=154$. Taking $\pi=\frac{22}{7}$, $r^2=154\cdot\frac{7}{22}=49$, so $r=7$. The circle is $(x-1)^2+(y+1)^2=49$.

Question 24

Find the equation of the circle which passes through the points $(2,3)$ and $(4,5)$ and the centre lies on the straight line $y-4x+3=0$.

Solution. The centre is equidistant from the two points, so it lies on their perpendicular bisector. The chord has midpoint $(3,4)$ and slope $1$, so the bisector has slope $-1$: $y-4=-(x-3)$, that is $x+y=7$. The centre also lies on $y=4x-3$. Substituting, $x+4x-3=7$, so $x=2$, $y=5$. Centre $(2,5)$.

The radius squared is the squared distance to $(2,3)$: $0+4=4$. The circle is $(x-2)^2+(y-5)^2=4$.

Question 25

Find the equation of a circle whose centre is $(3,-1)$ and which cuts off a chord of length $6$ units on the line $2x-5y+18=0$.

Solution. The perpendicular from the centre to a chord bisects it, so the radius, the distance $d$ to the line and the half-chord $3$ form a right triangle: $r^2=d^2+3^2$. The distance is

$$d=\frac{|2(3)-5(-1)+18|}{\sqrt{4+25}}=\frac{29}{\sqrt{29}}=\sqrt{29}.$$

So $r^2=29+9=38$ and the circle is $(x-3)^2+(y+1)^2=38$.

Question 26

Find the equation of a circle of radius $5$ which is touching another circle $x^2+y^2-2x-4y-20=0$ at $(5,5)$.

Solution. The given circle has centre $(1,2)$ and $r^2=1+4+20=25$, so radius $5$; the point $(5,5)$ is on it ($25+25-10-20-20=0$). Two circles that touch at a point have their centres and the point of contact on one line. The line from $(1,2)$ to $(5,5)$ has direction $(4,3)$, of length $5$.

The new centre is at distance $5$ from $(5,5)$ along this line, so it is $(5,5)\pm(4,3)$. The choice with the minus sign returns $(1,2)$, which is just the given circle again and does not count as a different circle touching it. The plus sign gives the centre $(9,8)$, the circle touching externally: $(x-9)^2+(y-8)^2=25$.

Question 27

Find the equation of a circle passing through the point $(7,3)$ having radius $3$ units and whose centre lies on the line $y=x-1$.

Solution. The centre is $(h,h-1)$. Its distance from $(7,3)$ must be $3$:

$$(h-7)^2+(h-4)^2=9\ \Rightarrow\ 2h^2-22h+56=0\ \Rightarrow\ h^2-11h+28=0\ \Rightarrow\ h=4\text{ or }7.$$

Both roots are valid, so there are two circles: centre $(4,3)$ and centre $(7,6)$, each with radius $3$. The printed key lists only the first. A complete answer gives both.

Question 28

Find the equation of each of the following parabolas.

Question 28 (a)

Directrix $x=0$, focus at $(6,0)$.

Solution. The vertex is halfway between focus and directrix, at $(3,0)$, so $a=3$ and the parabola opens to the right: $y^2=4a(x-3)=12(x-3)$.

Question 28 (b)

Vertex at $(0,4)$, focus at $(0,2)$.

Solution. The same data as Question 18: $a=2$, opening downward, $x^2=-8(y-4)$.

Question 28 (c)

Focus at $(-1,-2)$, directrix $x-2y+3=0$.

Solution. The directrix is not parallel to an axis, so use the definition directly: distance to the focus equals distance to the line, and squaring avoids the square root:

$$(x+1)^2+(y+2)^2=\frac{(x-2y+3)^2}{5}.$$

Multiply by $5$ and expand: $5x^2+5y^2+10x+20y+25=x^2+4y^2-4xy+6x-12y+9$, which collects to the equation below. The $xy$ term shows the axis is tilted.

Question 29

Find the equation of the set of all points the sum of whose distances from the points $(3,0)$ and $(9,0)$ is $12$.

Solution. A constant sum of distances to two fixed points, larger than the distance between them ($6$), is an ellipse with those points as foci. The sum is $2a=12$, so $a=6$; the foci are $6$ apart, so $c=3$ and $b^2=36-9=27$. The centre is the midpoint $(6,0)$:

$$\frac{(x-6)^2}{36}+\frac{y^2}{27}=1\ \Rightarrow\ 3(x-6)^2+4y^2=108,$$

which expands to the form below.

Question 30

Find the equation of the set of all points whose distance from $(0,4)$ is $\dfrac23$ of their distance from the line $y=9$.

Solution. Distance from $(0,4)$ is $\sqrt{x^2+(y-4)^2}$ and from the line is $|y-9|$. Squaring,

$$9\left[x^2+(y-4)^2\right]=4(y-9)^2\ \Rightarrow\ 9x^2+9y^2-72y+144=4y^2-72y+324,$$

so $9x^2+5y^2=180$. The ratio $\frac23<1$ confirms an ellipse.

Question 31

Show that the set of all points such that the difference of their distances from $(4,0)$ and $(-4,0)$ is always equal to $2$ represents a hyperbola.

Solution. Let $d_1=\sqrt{(x+4)^2+y^2}$ and $d_2=\sqrt{(x-4)^2+y^2}$ and take $d_1-d_2=2$ (the other branch is $d_2-d_1=2$). The foci are $8$ apart, more than $2$, so a real locus exists. Rather than squaring twice, use $d_1^2-d_2^2=16x$. Then $(d_1-d_2)(d_1+d_2)=16x$ gives $d_1+d_2=8x$, and together with $d_1-d_2=2$,

$$d_1=4x+1,\qquad d_2=4x-1.$$

Squaring $d_2$: $(x-4)^2+y^2=(4x-1)^2$, so $x^2-8x+16+y^2=16x^2-8x+1$, which gives $15x^2-y^2=15$:

$$x^2-\frac{y^2}{15}=1.$$

This is the standard hyperbola with $a=1$, $b^2=15$, $c^2=16$, so its foci are indeed $(\pm4,0)$.

Question 32

Find the equation of the hyperbola with

Question 32 (a)

Vertices $(\pm5,0)$, foci $(\pm7,0)$.

Solution. The vertices give $a=5$ and the foci give $c=7$, so $b^2=c^2-a^2=49-25=24$. The foci lie on the $x$-axis:

$$\frac{x^2}{25}-\frac{y^2}{24}=1.$$

The printed key gives $15x^2-y^2=15$, which is wrong. That is $x^2-\frac{y^2}{15}=1$, the hyperbola of Question 31, with vertices at $(\pm1,0)$ and foci at $(\pm4,0)$. It has been copied across from the wrong question.

Question 32 (b)

Vertices $(0,\pm7)$, $e=\dfrac43$.

Solution. The vertices are on the $y$-axis, so the transverse axis is vertical with $a=7$. Then $c=ae=\frac{28}{3}$, and $b^2=c^2-a^2=\frac{784}{9}-49=\frac{343}{9}$:

$$\frac{y^2}{49}-\frac{9x^2}{343}=1\ \Rightarrow\ 7y^2-9x^2=343.$$

Question 32 (c)

Foci $(0,\pm\sqrt{10})$, passing through $(2,3)$.

Solution. Foci on the $y$-axis give $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ with $a^2+b^2=10$. Passing through $(2,3)$: $\frac{9}{a^2}-\frac{4}{b^2}=1$. Put $A=a^2$, so $b^2=10-A$:

$$9(10-A)-4A=A(10-A)\ \Rightarrow\ A^2-23A+90=0\ \Rightarrow\ A=5\text{ or }18.$$

Since $b^2=10-A$ must be positive, $A<10$, so $A=5$ and $b^2=5$. The hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$.

State whether the statements in each of Questions 33 to 40 are True or False. Justify.

Question 33

The line $x+3y=0$ is a diameter of the circle $x^2+y^2+6x+2y=0$.

Solution. A line is a diameter exactly when it passes through the centre. The centre is $(-3,-1)$, and substituting gives $-3+3(-1)=-6\neq0$. The line misses the centre, so the statement is false. (The line $x-3y=0$ would work: $-3+3=0$.)

Question 34

The shortest distance from the point $(2,-7)$ to the circle $x^2+y^2-14x-10y-151=0$ is equal to $5$.

Solution. The circle has centre $(7,5)$ and $r^2=49+25+151=225$, so $r=15$. The distance from $(2,-7)$ to the centre is $\sqrt{25+144}=13$, which is less than $15$, so the point is inside the circle. The shortest distance to the circle is then $r-13=2$ (the hint’s “difference of radius and distance”), not $5$.

Question 35

If the line $lx+my=1$ is a tangent to the circle $x^2+y^2=a^2$, then the point $(l,m)$ lies on a circle.

Solution. Tangency means the distance from the origin to the line equals the radius: $\frac{1}{\sqrt{l^2+m^2}}=a$, so $l^2+m^2=\frac{1}{a^2}$. This is the equation of a circle in the $(l,m)$-plane, centred at the origin with radius $\frac1a$.

Question 36

The point $(1,2)$ lies inside the circle $x^2+y^2-2x+6y+1=0$.

Solution. The circle has centre $(1,-3)$ and $r^2=1+9-1=9$, so $r=3$. The distance from $(1,2)$ to the centre is $5>3$, so the point is outside. (Equivalently, the left side at $(1,2)$ is $1+4-2+12+1=16>0$, and a positive value means outside.)

Question 37

The line $lx+my+n=0$ will touch the parabola $y^2=4ax$ if $ln=am^2$.

Solution. From the line, $x=-\frac{my+n}{l}$. Substituting in $y^2=4ax$ gives $ly^2+4amy+4an=0$. Tangency needs a zero discriminant: $16a^2m^2-16a\,ln=0$, so $am^2=ln$. This is the condition stated.

Question 38

If $P$ is a point on the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ whose foci are $S$ and $S’$, then $PS+PS’=8$.

Solution. For any point of an ellipse, $PS+PS’$ equals the length of the major axis. Here the larger denominator is under $y^2$, so $a^2=25$ and $a=5$ (the major axis is along the $y$-axis). The sum is $2a=10$, not $8$. The number $8$ would be correct only if the ellipse were read as having semi-major axis $4$, the mistake of picking the first denominator.

Question 39

The line $2x+3y=12$ touches the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=2$ at the point $(3,2)$.

Solution. First, $(3,2)$ is on the ellipse: $\frac99+\frac44=2$. The tangent at $(x_1,y_1)$ to $\frac{x^2}{9}+\frac{y^2}{4}=2$ is $\frac{xx_1}{9}+\frac{yy_1}{4}=2$, which at $(3,2)$ is $\frac x3+\frac y2=2$, i.e. $2x+3y=12$. This is the given line. (The right-hand side being $2$, not $1$, does not change the tangent rule.)

Question 40

The locus of the point of intersection of the lines $\sqrt3x-y-4\sqrt3k=0$ and $\sqrt3kx+ky-4\sqrt3=0$ for different values of $k$ is a hyperbola whose eccentricity is $2$.

Solution. Eliminate $k$. The first line gives $\sqrt3x-y=4\sqrt3\,k$. The second gives $k(\sqrt3x+y)=4\sqrt3$. Multiplying the two,

$$(\sqrt3x-y)\cdot k(\sqrt3x+y)=4\sqrt3k\cdot4\sqrt3\ \Rightarrow\ 3x^2-y^2=48\quad(k\ne0),$$

so $\frac{x^2}{16}-\frac{y^2}{48}=1$. This is a hyperbola with $a^2=16$, $b^2=48$, so $e^2=1+\frac{48}{16}=4$ and $e=2$.

Common mistakes

• Q23: Using $\pi=3.14$ and getting an irrational radius. The textbook value $154$ is built on $\pi=\frac{22}{7}$, which makes $r=7$ exactly.
• Q26: Stopping at the centre $(1,2)$ because it is the first thing the line through $(5,5)$ gives. That is the given circle itself, not a second circle touching it.
• Q27: Keeping only one root of $h^2-11h+28=0$. A centre on a line at a fixed distance from a point can be on either side, so two circles exist.
• Q30, Q31: Squaring without moving a root first, or forgetting that a distance to a line is $|y-9|$ and squaring removes the sign. In Question 31, trying to square the difference twice rather than using $d_1^2-d_2^2=16x$ makes the algebra twice as long.
• Q32(a): Copying the form $15x^2-y^2=15$ from Question 31. The vertices $(\pm5,0)$ force $a^2=25$; always check the vertex of your final answer.
• Q34, Q38: Taking the shortest distance as $13+15$ or $13-5$ without first checking whether the point is inside; and reading the ellipse’s $a$ from the first denominator instead of the larger one.

Practise next

• Other pages in this unit: Short Answer Type, Q1–22 and Fill in the Blanks and Objective Type, Q41–59.
• NCERT textbook: Exercise 10.1 for circle problems, Exercise 10.3 for ellipses, Exercise 10.4 for hyperbolas and the Miscellaneous Exercise.