Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 1.

Find the

(i) lengths of the principal axes

(ii) co-ordinates of the foci

(iii) equations of directrices

(iv) length of the latus rectum

(v) distance between foci

(vi) distance between directrices of the ellipse:

(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

(b) 3x^{2} + 4y^{2} = 12

(c) 2x^{2} + 6y^{2} = 6

(d) 3x^{2} + 4y^{2} = 1

Solution:

(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 25 and b^{2} = 9

a = 5 and b = 3

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10

Length of minor axis = 2b = 2(3) = 6

Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)

= \(\frac{\sqrt{25-9}}{5}\)

= \(\frac{\sqrt{16}}{5}\)

= \(\frac{4}{5}\)

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),

i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0)

i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)

= \(\pm \frac{5}{\frac{4}{5}}\)

= \(\pm \frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae

= 2(5)(\(\frac{4}{5}\))

= 8

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)

= \(\frac{2(5)}{\frac{4}{5}}\)

= \(\frac{25}{2}\)

(b) Given equation of the ellipse is 3x^{2} + 4y^{2} = 12

\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 4 and b^{2} = 3

a = 2 and b = √3

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(2) = 4

Length of minor axis = 2b = 2√3

Lengths of the principal axes are 4 and 2√3.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)

= \(\frac{\sqrt{4-3}}{2}\)

= \(\frac{1}{2}\)

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),

i.e., S(2(\(\frac{1}{2}\)), 0) and S'(-2(\(\frac{1}{2}\)), 0)

i.e., S(1, 0) and S'(-1, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)

= \(\pm \frac{2}{\frac{1}{2}}\)

= ±4

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(\sqrt{3})^{2}}{2}=3\)

(v) Distance between foci = 2ae = 2(2)(\(\frac{1}{2}\)) = 2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)

= \(\frac{2(2)}{\frac{1}{2}}\)

= 8

(c) Given equation of the ellipse is 2x^{2} + 6y^{2} = 6

\(\frac{x^{2}}{3}+\frac{y^{2}}{1}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 3 and b^{2} = 1

a = √3 and b = 1

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3

Length of minor axis = 2b = 2(1) = 2

Lengths of the principal axes are 2√3 and 2.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)

= \(\frac{\sqrt{3-1}}{\sqrt{3}}\)

= \(\frac{\sqrt{2}}{\sqrt{3}}\)

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),

i.e., S(√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), o) and S'(-√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), 0)

i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),

= \(\pm \frac{\sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)

= \(\pm \frac{3}{\sqrt{2}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(1)^{2}}{\sqrt{3}}=\frac{2}{\sqrt{3}}\)

(v) Distance between foci = 2ae

= \(2(\sqrt{3})\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\)

= 2√2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)

= \(\frac{2 \sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)

= \(\frac{2 \times 3}{\sqrt{2}}\)

= 3√2

(d) Given equation of the ellipse is 3x^{2} + 4y = 1.

\(\frac{x^{2}}{\frac{1}{3}}+\frac{y^{2}}{\frac{1}{4}}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = \(\frac{1}{3}\) and b^{2} = \(\frac{1}{4}\)

a = \(\frac{1}{\sqrt{3}}\) and b = \(\frac{1}{2}\)

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(\(\frac{1}{\sqrt{3}}\)) = \(\frac{2}{\sqrt{3}}\)

Length of minor axis = 2b = 2(\(\frac{1}{2}\)) = 1

Lengths of the principal axes are \(\frac{2}{\sqrt{3}}\) and 1.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)

e = \(\frac{\sqrt{\frac{1}{3}-\frac{1}{4}}}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{\frac{1}{12}}}{\frac{1}{\sqrt{3}}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),

i.e., S\(\left(\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\) and S’\(\left(-\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\)

i.e., S(\(\frac{1}{2 \sqrt{3}}\), 0) and S'(-\(\frac{1}{2 \sqrt{3}}\), 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),

= \(\pm \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2}}\)

= \(\pm \frac{2}{\sqrt{3}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\)

= \(\frac{2\left(\frac{1}{2}\right)^{2}}{\frac{1}{\sqrt{3}}}\)

= \(\frac{\sqrt{3}}{2}\)

(v) Distance between foci = 2ae

= \(2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right)\)

= \(\frac{1}{\sqrt{3}}\)

(vi) Distance between directrices = \(\frac{2 a}{e}\)

= \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{2}}\)

= \(\frac{4}{\sqrt{3}}\)

Question 2.

Find the equation of the ellipse in standard form if

(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.

(ii) the length of the major axis is 10 and the distance between foci is 8.

(iii) distance between directrices is 18 and eccentricity is \(\frac{1}{3}\).

(iv) minor axis is 16 and eccentricity is \(\frac{1}{3}\).

(v) the distance between foci is 6 and the distance between directrices is \(\frac{50}{3}\).

(vi) the latus rectum has length 6 and foci are (±2, 0).

(vii) passing through the points (-3, 1) and (2, -2).

(viii) the distance between its directrices is 10 and which passes through (-√5, 2).

(ix) eccentricity is \(\frac{2}{3}\) and passes through (2, \(\frac{-5}{3}\)).

Solution:

(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Given, eccentricity (e) = \(\frac{3}{8}\)

Distance between foci = 2ae

Given, distance between foci = 6

2ae = 6

The required equation of ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\).

(ii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Length of major axis = 2a

Given, length of major axis = 10

2a = 10

a = 5

a^{2} = 25

Distance between foci = 2ae

Given, distance between foci = 8

2ae = 8

2(5)e = 8

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Given, eccentricity (e) = \(\frac{1}{3}\)

Distance between directrices = \(\frac{2a}{e}\)

Given, distance between directrices = 18

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{8}=1\)

(iv) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Length of minor axis = 2b

Given, length of minor axis = 16

2b = 16

b = 8

b^{2} = 64

Given, eccentricity (e) = \(\frac{1}{3}\)

Now, b^{2} = a^{2} (1 – e^{2})

The required equation of ellipse is \(\frac{x^{2}}{72}+\frac{y^{2}}{64}=1\).

(v) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Distance between foci = 2ae

Given, distance between foci = 6

2ae = 6

ae = 3

a = \(\frac{3}{e}\) …….(i)

Distance between directrices = \(\frac{2a}{e}\)

Given, distance between directrices = \(\frac{50}{3}\)

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

(vi) Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0).

The foci of the ellipse are on the X-axis.

Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Length of latus rectum = \(\frac{2 b^{2}}{a}\)

\(\frac{2 b^{2}}{a}\) = 6

b^{2} = 3a …..(i)

Co-ordinates of foci are (±ae, 0)

ae = 2

a^{2}e^{2} = 4 …..(ii)

Now, b^{2} = a^{2} (1 – e^{2})

b^{2} = a^{2} – a^{2} e^{2}

3a = a^{2} – 4 …..[From (i) and (ii)]

a^{2} – 3a – 4 = 0

a^{2} – 4a + a – 4 = 0

a(a – 4) + 1(a – 4) = 0

(a – 4) (a + 1) = 0

a – 4 = 0 or a + 1 = 0

a = 4 or a = -1

Since a = -1 is not possible,

a = 4

a^{2} = 16

Substituting a = 4 in (i), we get

b^{2} = 3(4) = 12

The required equation of ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\).

(vii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

The ellipse passes through the points (-3, 1) and (2, -2).

Substituting x = -3 and y = 1 in equation of ellipse, we get

Equations (i) and (ii) become

9A + B = 1 …..(iii)

4A + 4B = 1 …..(iv)

Multiplying (iii) by 4, we get

36A + 4B = 4 …..(v)

Subtracting (iv) from (v), we get

32A = 3

A = \(\frac{3}{32}\)

Substituting A = \(\frac{3}{32}\) in (iv), we get

4(\(\frac{3}{32}\)) + 4B = 1

\(\frac{3}{8}\) + 4B = 1

4B = 1 – \(\frac{3}{8}\)

4B = \(\frac{5}{8}\)

B = \(\frac{5}{32}\)

(viii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Distance between directrices = \(\frac{2 a}{e}\)

Given, distance between directrices = 10

\(\frac{2 a}{e}\) = 10

a = 5e …..(i)

The ellipse passes through (-√5, 2).

Substituting x = -√5 and y = 2 in equation of ellipse, we get

b^{2} = 15(\(\frac{2}{5}\))

b^{2} = 6

The required equation of ellipse is \(\frac{x^{2}}{15}+\frac{y^{2}}{6}=1\).

(ix) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Given, eccentricity (e) = \(\frac{2}{3}\)

The ellipse passes through (2, \(\frac{-5}{3}\)).

Substituting x = 2 and y = \(\frac{-5}{3}\) in equation of ellipse, we get

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\).

Question 3.

Find the eccentricity of an ellipse, if the length of its latus rectum is one-third of its minor axis.

Solution:

Let the equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.

Length of latus rectum = \(\frac{2 b^{2}}{a}\)

Length of minor axis = 2b

According to the given condition,

Length of latus rectum = \(\frac{1}{3}\) (Minor axis)

Question 4.

Find the eccentricity of an ellipse, if the distance between its directrices is three times the distance between its foci.

Solution:

Let the required equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.

Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)

Distance between foci = 2ae

According to the given condition,

distance between directrices = 3(distance between foci)

\(\frac{2 \mathrm{a}}{\mathrm{e}}\) = 3(2ae)

\(\frac{1}{\mathrm{e}}\) = 3e

\(\frac{1}{3}\) = e^{2}

e = \(\frac{1}{\sqrt{3}}\) ……[∵ 0 < e < 1]

Eccentricity of the ellipse is \(\frac{1}{\sqrt{3}}\)

Question 5.

Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.

Solution:

Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 25, b^{2} = 16

a = 5, b = 4

Question 6.

A tangent having slope \(\left(-\frac{1}{2}\right)\) the ellipse 3x^{2} + 4y^{2} = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.

Solution:

Given equation of the ellipse is 3x^{2} + 4y^{2} = 12.

\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 4, b^{2} = 3

Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

Here, m = \(-\frac{1}{2}\)

Equations of the tangents are

y = \(\frac{-1}{2} x \pm \sqrt{4\left(-\frac{1}{2}\right)^{2}+3}=\frac{-1}{2} x \pm 2\)

2y = -x ± 4

x + 2y ± 4 = 0

Consider the tangent x + 2y – 4 = 0

Let this tangent intersect the X-axis at A(x_{1}, 0) and Y-axis at B(0, y_{1}).

x_{1} + 0 – 4 = 0 and 0 + 2y_{1} – 4 = 0

x_{1} = 4 and y_{1} = 2

A = (4, 0) and B = (0, 2)

l(OA) = 4 and l(OB) = 2

Area of ∆AOB = \(\frac{1}{2}\) × l(OA) × l(OB)

= \(\frac{1}{2}\) × 4 × 2

= 4 sq.units

Question 7.

Show that the line x – y = 5 is a tangent to the ellipse 9x^{2} + 16y^{2} = 144. Find the point of contact.

Solution:

Given equation of the ellipse is 9x^{2} + 16y^{2} = 144

\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 16 and b^{2} = 9

Given equation of line is x – y = 5, i.e., y = x – 5

c^{2} = a^{2} m^{2} + b^{2}

Comparing this equation with y = mx + c, we get

m = 1 and c = -5

For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have

c^{2} = a^{2} m^{2} + b^{2}

c^{2} = (-5)^{2} = 25

a^{2} m^{2} + b^{2} = 16(1)^{2} + 9 = 16 + 9 = 25 = c^{2}

The given line is a tangent to the given ellipse and point of contact

= \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)

= \(\left(\frac{(-16)(1)}{-5}, \frac{9}{-5}\right)\)

= \(\left(\frac{16}{5}, \frac{-9}{5}\right)\)

Question 8.

Show that the line 8y + x = 17 touches the ellipse x^{2} + 4y^{2} = 17. Find the point of contact.

Solution:

Given equation of the ellipse is x^{2} + 4y^{2} = 17.

\(\frac{x^{2}}{17}+\frac{y^{2}}{\frac{17}{4}}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 17 and b^{2} = \(\frac{17}{4}\)

Given equation of line is 8y + x = 17,

y = \(\frac{-1}{8} x+\frac{17}{8}\)

Comparing this equation with y = mx + c, we get

m = \(\frac{-1}{8}\) and c = \(\frac{17}{8}\)

For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have

Question 9.

Determine whether the line x + 3y√2 = 9 is a tangent to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). If so, find the co-ordinates of the point of contact.

Solution:

Given equation of the ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 9 and b^{2} = 4

Given equation of line is x + 3y√2 = 9,

i.e., y = \(\frac{-1}{3 \sqrt{2}} x+\frac{3}{\sqrt{2}}\)

Comparing this equation with y = mx + c, we get

m = \(\frac{-1}{3 \sqrt{2}}\) and c = \(\frac{3}{\sqrt{2}}\)

For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=\)1, we must have

Question 10.

Find k, if the line 3x + 4y + k = 0 touches 9x^{2} + 16y^{2} = 144.

Solution:

Given equation of the ellipse is 9x^{2} + 16y^{2 }= 144.

\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 16 and b^{2} = 9

Given equation of line is 3x + 4y + k = 0,

i.e., y = \(-\frac{3}{4} x-\frac{k}{4}\)

Comparing this equation with y = mx + c, we get

m = \(\frac{-3}{4}\) and c = \(\frac{-\mathrm{k}}{4}\)

For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have

c^{2} = a^{2} m^{2} + b^{2}

\(\left(\frac{-k}{4}\right)^{2}=16\left(\frac{-3}{4}\right)^{2}+9\)

\(\frac{\mathrm{k}^{2}}{16}\) = 9 + 9

\(\frac{\mathrm{k}^{2}}{16}\) = 18

k^{2} = 288

k = ±12√2

Question 11.

Find the equations of the tangents to the ellipse:

(i) \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).

(ii) 4x^{2} + 7y^{2} = 28 from the point (3, -2).

(iii) 2x^{2} + y^{2} = 6 from the point (2, 1).

(iv) x^{2} + 4y^{2} = 9 which are parallel to the line 2x + 3y – 5 = 0.

(v) \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\) which are parallel to the line x + y + 1 = 0.

(vi) 5x^{2} + 9y^{2} = 45 which are ⊥ to the line 3x + 2y + 1 = 0.

(vii) x^{2} + 4y^{2} = 20 which are ⊥ to the line 4x + 3y = 7.

Solution:

(i) Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\).

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 5 and b^{2} = 4

Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

Since (2, -2) lies on both the tangents,

-2 = 2m ±\(\sqrt{5 m^{2}+4}\)

-2 – 2m = ±\(\sqrt{5 m^{2}+4}\)

Squaring both the sides, we get

4m^{2} + 8m + 4 = 5m^{2} + 4

m^{2} – 8m = 0

m(m – 8) = 0

m = 0 or m = 8

These are the slopes of the required tangents.

By slope point form y – y_{1} = m(x – x_{1}),

the equations of the tangents are

y + 2 = 0(x – 2) and y + 2 = 8(x – 2)

y + 2 = 0 and y + 2 = 8x – 16

y + 2 = 0 and 8x – y – 18 = 0

(ii) Given equation of the ellipse is 4x^{2} + 7y^{2} = 28.

\(\frac{x^{2}}{7}+\frac{y^{2}}{4}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 7 and b^{2} = 4

Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

Since (3, -2) lies on both the tangents,

-2 = 3m ± \(\sqrt{7 \mathrm{~m}^{2}+4}\)

-2 – 3m = ±\(\sqrt{7 \mathrm{~m}^{2}+4}\)

Squaring both the sides, we get

9m^{2} + 12m + 4 = 7m^{2} + 4

2m^{2} + 12m = 0

2m(m + 6) = 0

m = 0 or m = -6

These are the slopes of the required tangents.

By slope point form y – y_{1} = m(x – x_{1}),

the equations of the tangents are

y + 2 = 0(x – 3) and y + 2 = -6(x – 3)

y + 2 = 0 and y + 2 = -6x + 18

y + 2 = 0 and 6x + y – 16 = 0

(iii) Given equation of the ellipse is 2x^{2} + y^{2} = 6.

\(\frac{x^{2}}{3}+\frac{y^{2}}{6}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 3 and b^{2} = 6

Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

Since (2, 1) lies on both the tangents,

1 = 2m ± \(\sqrt{3 m^{2}+6}\)

1 – 2m = ±\(\sqrt{3 m^{2}+6}\)

Squaring both the sides, we get

1 – 4m + 4m^{2} = 3m^{2} + 6

m^{2} – 4m – 5 = 0

(m – 5)(m + 1) = 0

m = 5 or m = -1

These are the slopes of the required tangents.

By slope point form y – y_{1} = m(x – x_{1}),

the equations of the tangents are

y – 1 = 5(x – 2) and y – 1 = -1(x – 2)

y – 1 = 5x – 10 and y – 1 = -x + 2

5x – y – 9 = 0 and x + y – 3 = 0

(iv) Given equation of the ellipse is x^{2} + 4y^{2} = 9.

\(\frac{x^{2}}{9}+\frac{y^{2}}{\frac{9}{4}}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 9 and b^{2} = \(\frac{9}{4}\)

Slope of the line 2x + 3y – 5 = 0 is \(\frac{-2}{3}\).

Since the given line is parallel to the required tangents, slope of the required tangents is

m = \(\frac{-2}{3}\)

Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

(v) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\).

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 25 and b^{2} = 4

Slope of the given line x + y + 1 = 0 is -1.

Since the given line is parallel to the required tangents,

the slope of the required tangents is m = -1.

Equations of tangents to the ellipse

(vi) Given equation of the ellipse is 5x^{2} + 9y^{2} = 45.

\(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 9 and b^{2} = 5

Slope of the given line 3x + 2y + 1 = 0 is \(\frac{-3}{2}\)

Since the given line is perpendicular to the required tangents, slope of the required tangents is

m = \(\frac{2}{3}\)

Equations of tangents to the ellipse

(vii) Given equation of the ellipse is x^{2} + 4y^{2} = 20.

\(\frac{x^{2}}{20}+\frac{y^{2}}{5}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 20 and b^{2} = 5

Slope of the given line 4x + 3y = 7 is \(\frac{-4}{3}\).

Since the given line is perpendicular to the required tangents,

slope of the required tangents is m = \(\frac{3}{4}\).

Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

y = \(\frac{3}{4} x \pm \sqrt{20\left(\frac{3}{4}\right)^{2}+5}\)

Question 12.

Find the equation of the locus of a point, the tangents from which to the ellipse 3x^{2} + 5y^{2} = 15 are at right angles.

Solution:

Given equation of the ellipse is 3x^{2} + 5y^{2} = 15.

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

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 5 and b^{2} = 3

Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

y = mx ± \(\sqrt{5 m^{2}+3}\)

y – mx =±\(\sqrt{5 m^{2}+3}\)

Squaring both the sides, we get

y^{2} – 2mxy + m^{2}x^{2} = 5m^{2} + 3

(x^{2} – 5) m^{2} – 2xym + (y^{2} – 3) = 0

The roots m_{1} and m_{2} of this quadratic equation are the slopes of the tangents.

m_{1}m_{2} = \(\frac{y^{2}-3}{x^{2}-5}\)

Since the tangents are at right angles,

m_{1}m_{2} = -1

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

y^{2} – 3 = -x^{2} + 5

x^{2} + y^{2} = 8, which is the required equation of the locus.

Alternate method:

The locus of the point of intersection of perpendicular tangents is the director circle of an ellipse.

The equation of the director circle of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is x^{2} + y^{2} = a^{2} + b^{2}

Here, a^{2} = 5 and b^{2} = 3

x^{2} + y^{2} = 5 + 3

x^{2} + y^{2} = 8, which is the required equation of the locus.

Question 13.

Tangents are drawn through a point P to the ellipse 4x^{2} + 5y^{2} = 20 having inclinations θ_{1} and θ_{2} such that tan θ_{1} + tan θ_{2} = 2. Find the equation of the locus of P.

Solution:

Given equation of the ellipse is 4x^{2} + 5y^{2} = 20.

\(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 5 and b^{2} = 4

Since inclinations of tangents are θ_{1} and θ_{2},

m_{1} = tan θ_{1} and m_{2} = tan θ_{2}

Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

y = mx ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)

y – mx = ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)

Squaring both the sides, we get

y^{2} – 2mxy + m^{2}x^{2} = 5m^{2} + 4

(x^{2} – 5)m^{2} – 2xym + (y^{2} – 4) = 0

The roots m_{1} and m_{2} of this quadratic equation are the slopes of the tangents.

m_{1} + m_{2} = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)

Given, tan θ_{1} + tan θ_{2} = 2

m_{1} + m_{2} = 2

\(\frac{2 x y}{x^{2}-5}\) = 2

xy = x^{2} – 5

x^{2} – xy – 5 = 0, which is the required equation of the locus of P.

Question 14.

Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse.

Solution:

Let P(θ_{1}) and Q(θ_{2}) be any two points on the given ellipse such that θ_{1} – θ_{2} = k, where k is a constant.

The equation of the tangent at point P(θ_{1}) is

\(\frac{x \cos \theta_{1}}{\mathrm{a}}+\frac{y \sin \theta_{1}}{\mathrm{~b}}=1\) ……(i)

The equation of the tangent at point Q(θ_{2}) is

\(\frac{x \cos \theta_{2}}{\mathrm{a}}+\frac{y \sin \theta_{2}}{\mathrm{~b}}=1\) ……(ii)

Multiplying equation (i) by cos θ_{2} and equation (ii) by cos θ_{1} and subtracting, we get

\(\frac{y}{b}\) (sin θ_{1} cos θ_{2} – sin θ_{2} cos θ_{1}) = cos θ_{2} – cos θ_{1}

Question 15.

P and Q are two points on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with eccentric angles θ_{1} and θ_{2}. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ_{1} + θ_{2} = \(\frac{\pi}{2}\).

Solution:

Given equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).

θ_{1} and θ_{2} are the eccentric angles of a tangent.

Equation of tangent at point P is

\(\frac{x}{a} \cos \theta_{1}+\frac{y}{b} \sin \theta_{1}=1\) ……(i)

Equation of tangent at point Q is

\(\frac{x}{a} \cos \theta_{2}+\frac{y}{b} \sin \theta_{2}=1\) ………(ii)

θ_{1} + θ_{2} = \(\frac{\pi}{2}\) …..[Given]

i.e., points P and Q coincide, which is not possible, as P and Q are two different points.

cos θ_{1} – sin θ_{1} ≠ 0

Dividing equation (iii) by (cos θ_{1} – sin θ_{1}), we get

\(\frac{x_{1}}{a}=\frac{y_{1}}{b}\)

bx_{1} – ay_{1} = 0

bx – ay = 0, which is the required equation of locus of point M.

Question 16.

The eccentric angles of two points P and Q of the ellipse 4x^{2} + y^{2} = 4 differ by \(\frac{2 \pi}{3}\). Show that the locus of the point of intersection of the tangents at P and Q is the ellipse 4x^{2} + y^{2} = 16.

Solution:

Given equation of the ellipse is 4x^{2} + y^{2} = 4.

\(\frac{x^{2}}{1}+\frac{y^{2}}{4}=1\)

Let P(θ_{1}) and Q(θ_{2}) be any two points on the given ellipse such that

θ_{1} – θ_{2} = \(\frac{2 \pi}{3}\)

The equation of the tangent at point P(θ_{1}) is

\(\frac{x \cos \theta_{1}}{1}+\frac{y \sin \theta_{1}}{2}=1\) ……(i)

The equation of the tangent at point Q(θ_{2}) is

\(\frac{x \cos \theta_{2}}{1}+\frac{y \sin \theta_{2}}{2}=1\)

Multiplying equation (i) by cos θ_{2} and equation (ii) by cos θ_{1} and subtracting, we get

Question 17.

Find the equations of the tangents to the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), making equal intercepts on co-ordinate axes.

Solution:

Given equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 16 and b^{2} = 9

Since the tangents make equal intercepts on the co-ordinate axes,

m = -1.

Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

y = -x ± \(\sqrt{16(-1)^{2}+9}\)

y = -x ± \(\sqrt{25}\)

x + y = ±5

Question 18.

A tangent having slope \(\left(-\frac{1}{2}\right)\) to the ellipse 3x^{2} + 4y^{2} = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.

Solution:

The equation of the ellipse is 3x^{2} + 4y^{2} = 12

\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)

Comparing with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 4, b^{2} = 3

The equation of tangent with slope m is

It meets X axis at A

∴ for A, put y = 0 in equation (1), we get,

x = ±4

∴ A = (±4, 0)

Similarly, B = (0, ±2)

∴ OA = 4, OB = 2

∴ Area of ΔOAB = \(\frac{1}{2}\) × OA × OB

= \(\frac{1}{2}\) × 4 × 2

= 4 sq. units