NCERT Exemplar Solutions Class 11 Maths Conic Sections — Short Answer Type Q1–22

NCERT Exemplar Problems, Class 11 Mathematics — Conic Sections, Short Answer Type questions 1 to 22. All 22 questions solved.

This band is a tour of all four conics. Questions 1 to 10 are about circles: centre-radius form $(x-h)^2+(y-k)^2=r^2$, the distance from a point to a line, and tangency as “distance from centre to line equals radius”. Questions 11 to 15 use the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $b^2=a^2(1-e^2)$, latus rectum $\frac{2b^2}{a}$ and directrices $x=\pm\frac ae$. Questions 16 to 19 are parabola work with $y^2=4ax$, and 20 to 22 are hyperbola work with $b^2=a^2(e^2-1)$.

Key insight. Every question here is decided by writing down what the conic is made of (centre and radius; $a$, $b$, $c$, $e$; focus and directrix) and then solving for the one missing number. Before you reach for a formula, ask which two of $a$, $b$, $c$, $e$ you have been given.

Question 1

Find the equation of the circle which touches both axes in the first quadrant and whose radius is $a$.

Solution. A circle that touches both axes has its centre equally far from each, and that distance is the radius. In the first quadrant the centre is therefore $(a, a)$, and the circle is $(x-a)^2+(y-a)^2=a^2$. Expanding and cancelling one $a^2$ gives the general form.

Question 2

Show that the point $(x, y)$ given by $x=\dfrac{2at}{1+t^2}$ and $y=\dfrac{a(1-t^2)}{1+t^2}$ lies on a circle for all real values of $t$ such that $-1\le t\le 1$, where $a$ is any given real number.

Solution. To show a point lies on a circle whatever $t$ is, eliminate $t$. Square both coordinates and add, because the numerators $2t$ and $1-t^2$ are the two legs of the identity $(2t)^2+(1-t^2)^2=(1+t^2)^2$:

$$x^2+y^2=\frac{4a^2t^2+a^2(1-t^2)^2}{(1+t^2)^2}=\frac{a^2(1+t^2)^2}{(1+t^2)^2}=a^2.$$

So every such point satisfies $x^2+y^2=a^2$, a circle with centre at the origin and radius $|a|$. The restriction $-1\le t\le1$ only decides which part of the circle is traced (for $a>0$ it is the upper half, $y\ge0$, since $1-t^2\ge0$); it does not change the circle.

Question 3

If a circle passes through the points $(0,0)$, $(a,0)$, $(0,b)$, then find the coordinates of its centre.

Solution. The three points form a right angle at the origin, and the angle in a semicircle is a right angle, so the hypotenuse joining $(a,0)$ and $(0,b)$ is a diameter. The centre is the midpoint of that diameter. (Equivalently, put $x^2+y^2+Dx+Ey=0$ through the points and get $D=-a$, $E=-b$, centre $\left(-\frac D2,-\frac E2\right)$.)

Question 4

Find the equation of the circle which touches the $x$-axis and whose centre is $(1, 2)$.

Solution. A circle touching the $x$-axis has radius equal to the distance of its centre from that axis, which is the $y$-coordinate, so $r=2$. Then $(x-1)^2+(y-2)^2=4$, which expands to the form below.

Question 5

If the lines $3x-4y+4=0$ and $6x-8y-7=0$ are tangents to a circle, then find the radius of the circle.

Solution. The lines are parallel, so the circle sits between them and the distance between them is a diameter. Divide the second line by $2$ so both have the same $x$ and $y$ coefficients: $3x-4y-\frac72=0$. The distance between $3x-4y+c_1=0$ and $3x-4y+c_2=0$ is $\frac{|c_1-c_2|}{\sqrt{3^2+4^2}}$:

$$d=\frac{\left|4-\left(-\frac72\right)\right|}{5}=\frac{15/2}{5}=\frac32.$$

The radius is half of this.

Question 6

Find the equation of a circle which touches both the axes and the line $3x-4y+8=0$ and lies in the third quadrant.

Solution. In the third quadrant a circle touching both axes with radius $a$ has centre $(-a,-a)$. It touches the line when the distance from the centre to the line equals $a$:

$$\frac{|-3a+4a+8|}{5}=a\ \Rightarrow\ |a+8|=5a.$$

Taking $a+8=5a$ gives $a=2$. The other sign, $a+8=-5a$, gives $a=-\frac43$, which is negative and so cannot be a radius. The circle is $(x+2)^2+(y+2)^2=4$.

Question 7

If one end of a diameter of the circle $x^2+y^2-4x-6y+11=0$ is $(3, 4)$, then find the coordinates of the other end of the diameter.

Solution. For $x^2+y^2+2gx+2fy+c=0$ the centre is $(-g,-f)$, so here the centre is $(2, 3)$. The centre is the midpoint of the diameter, so if the other end is $(p,q)$ then $\frac{3+p}2=2$ and $\frac{4+q}2=3$, giving $p=1$, $q=2$.

Question 8

Find the equation of the circle having $(1,-2)$ as its centre and passing through $3x+y=14$, $2x+5y=18$.

Solution. The wording means the circle passes through the point where these two lines meet. Eliminating $y$: multiply the first by $5$ to get $15x+5y=70$ and subtract the second, so $13x=52$, $x=4$, then $y=14-12=2$. The point is $(4,2)$.

The radius squared is the squared distance from the centre $(1,-2)$ to $(4,2)$: $3^2+4^2=25$. So $(x-1)^2+(y+2)^2=25$.

Question 9

If the line $y=\sqrt3\,x+k$ touches the circle $x^2+y^2=16$, then find the value of $k$.

Solution. Write the line as $\sqrt3x-y+k=0$. The circle has centre at the origin and radius $4$, so tangency requires the perpendicular distance from the origin to be $4$:

$$\frac{|k|}{\sqrt{3+1}}=4\ \Rightarrow\ |k|=8.$$

Both signs work, because there are two tangents of that slope, one on each side of the circle.

Question 10

Find the equation of a circle concentric with the circle $x^2+y^2-6x+12y+15=0$ and has double of its area.

Solution. Concentric means the same centre. For the given circle, centre $(3,-6)$ and $r^2=9+36-15=30$. Area is $\pi r^2$, so doubling the area doubles $r^2$ (not $r$): $r^2=60$. The new circle is $(x-3)^2+(y+6)^2=60$, which expands with constant term $9+36-60=-15$.

Question 11

If the latus rectum of an ellipse is equal to half of its minor axis, then find its eccentricity.

Solution. For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ the latus rectum is $\frac{2b^2}{a}$ and the minor axis is $2b$, so half of it is $b$. Setting $\frac{2b^2}{a}=b$ gives $a=2b$. Then

$$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac14}=\frac{\sqrt3}2.$$

Question 12

Given the ellipse with equation $9x^2+25y^2=225$, find the eccentricity and foci.

Solution. Divide by $225$ to reach standard form: $\frac{x^2}{25}+\frac{y^2}{9}=1$. So $a^2=25$ (larger, under $x^2$) and $b^2=9$, meaning the major axis lies along the $x$-axis. Then $c=\sqrt{a^2-b^2}=4$, $e=\frac ca=\frac45$ and the foci are $(\pm c,0)$.

Question 13

If the eccentricity of an ellipse is $\dfrac58$ and the distance between its foci is $10$, then find the latus rectum of the ellipse.

Solution. The distance between the foci is $2c=2ae=10$, so $ae=5$. With $e=\frac58$ this gives $a=8$. Then $b^2=a^2-c^2=64-25=39$, and the latus rectum is $\frac{2b^2}{a}=\frac{78}{8}$.

Question 14

Find the equation of ellipse whose eccentricity is $\dfrac23$, latus rectum is $5$ and the centre is $(0,0)$.

Solution. From $e=\frac23$: $b^2=a^2(1-e^2)=\frac59a^2$. The latus rectum gives $\frac{2b^2}{a}=5$, so $\frac{2}{a}\cdot\frac59a^2=\frac{10a}{9}=5$ and $a=\frac92$. Then $a^2=\frac{81}4$ and $b^2=\frac59\cdot\frac{81}{4}=\frac{45}4$. Taking the major axis along the $x$-axis, as the book does:

$$\frac{x^2}{81/4}+\frac{y^2}{45/4}=1.$$

(If the major axis were along the $y$-axis the roles of $x$ and $y$ would swap; the question does not say, and the book’s answer takes the $x$-axis.)

Question 15

Find the distance between the directrices of the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{20}=1$.

Solution. Here $a^2=36$, $b^2=20$, so $c^2=16$, $c=4$ and $e=\frac46=\frac23$. The directrices are $x=\pm\frac ae=\pm6\cdot\frac32=\pm9$, so the distance between them is $2\cdot9$. Equivalently $\frac{2a}{e}=\frac{2a^2}{c}=\frac{72}{4}$.

Question 16

Find the coordinates of a point on the parabola $y^2=8x$ whose focal distance is $4$.

Solution. Comparing with $y^2=4ax$ gives $a=2$. The focal distance of a point $(x,y)$ on the parabola equals its distance from the directrix $x=-2$, which is $x+2$. Setting $x+2=4$ gives $x=2$, then $y^2=16$, so $y=\pm4$.

Question 17

Find the length of the line-segment joining the vertex of the parabola $y^2=4ax$ and a point on the parabola where the line-segment makes an angle $\theta$ to the $x$-axis.

Solution. The vertex is the origin. A point on the parabola on a line through the origin at angle $\theta$ has $y=x\tan\theta$. Substituting, $x^2\tan^2\theta=4ax$, so $x=4a\cot^2\theta$ (ignoring the vertex itself, $x=0$). The segment has length $\frac{x}{\cos\theta}$:

$$\text{length}=\frac{4a\cos^2\theta}{\sin^2\theta\cos\theta}=\frac{4a\cos\theta}{\sin^2\theta}.$$

Question 18

If the points $(0,4)$ and $(0,2)$ are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Solution. The axis is the $y$-axis (both points have $x=0$). The distance from vertex to focus is $a=2$, and the focus lies below the vertex, so the parabola opens downward: $(x-0)^2=-4a(y-4)=-8(y-4)$. This gives $x^2=32-8y$.

Question 19

If the line $y=mx+1$ is tangent to the parabola $y^2=4x$ then find the value of $m$.

Solution. Substitute the line into the parabola: $(mx+1)^2=4x$, that is $m^2x^2+(2m-4)x+1=0$. A tangent meets the parabola at exactly one point, so the discriminant is zero:

$$(2m-4)^2-4m^2=0\ \Rightarrow\ 16-16m=0\ \Rightarrow\ m=1.$$

The line $y=1$ ($m=0$) also meets the parabola only once, but it is parallel to the axis rather than a tangent; the quadratic then drops to a linear equation and the discriminant test rightly does not select it.

Question 20

If the distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt2$, then obtain the equation of the hyperbola.

Solution. $2c=16$ so $c=8$. From $e=\frac ca=\sqrt2$, $a=\frac{8}{\sqrt2}$, so $a^2=32$. For a hyperbola $b^2=c^2-a^2=64-32=32$. Taking the foci on the $x$-axis (the question gives no axis, and the book assumes this): $\frac{x^2}{32}-\frac{y^2}{32}=1$, a rectangular hyperbola, because $e=\sqrt2$ always forces $a=b$.

Question 21

Find the eccentricity of the hyperbola $9y^2-4x^2=36$.

Solution. Divide by $36$: $\frac{y^2}{4}-\frac{x^2}{9}=1$. The $y^2$ term is positive, so the transverse axis is vertical, with $a^2=4$ (under the positive term) and $b^2=9$. Then $e^2=1+\frac{b^2}{a^2}=1+\frac94=\frac{13}4$.

Question 22

Find the equation of the hyperbola with eccentricity $\dfrac32$ and foci at $(\pm2,0)$.

Solution. The foci $(\pm2,0)$ give $c=2$, and $e=\frac ca=\frac32$ gives $a=\frac43$, $a^2=\frac{16}{9}$. Then $b^2=c^2-a^2=4-\frac{16}9=\frac{20}9$. The equation is

$$\frac{x^2}{16/9}-\frac{y^2}{20/9}=1\ \Longleftrightarrow\ \frac{9x^2}{16}-\frac{9y^2}{20}=1\ \Longleftrightarrow\ \frac{x^2}{4}-\frac{y^2}{5}=\frac49,$$

where the last form is what the book prints (multiply the middle form through by $\frac49$).

Common mistakes

• Q5: Using $6x-8y-7=0$ and $3x-4y+4=0$ directly in the parallel-line distance formula. The coefficients of $x$ and $y$ must match first, otherwise the numerator and denominator are on different scales. Also forgetting that the distance is the diameter, not the radius.
• Q6: Keeping the root $a=-\frac43$. A radius is positive, and the sign of the centre $(-a,-a)$ already carries the quadrant.
• Q10: Doubling the radius instead of the area. Area is proportional to $r^2$, so $r^2$ doubles, and $r$ goes up by $\sqrt2$ only.
• Q11, Q13: Using the full minor axis $2b$ instead of half of it, or confusing the distance between the foci ($2ae$) with $ae$.
• Q15: Finding the directrix $x=\frac{a}{e}$ and then reporting that as the distance. There are two directrices, so the gap is twice that.
• Q21: Treating $9y^2-4x^2=36$ as a hyperbola along the $x$-axis. When $y^2$ carries the plus sign, the denominator under $y^2$ is $a^2$.

Practise next

• The next pages in this unit: Long Answer Type and True or False, Q23–40 and Fill in the Blanks and Objective Type, Q41–59.
• NCERT textbook: Exercise 10.1 for circles, Exercise 10.2 for parabolas, Exercise 10.3 for ellipses and Exercise 10.4 for hyperbolas.