Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Ellipse Solutions Exercise 4(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Ellipse Solutions Exercise 4(b)

I.

Question 1.

Find the equation of tangent and normal to the ellipse x + 8y = 33 at (-1, 2).

Solution:

Equation of the tangent is

\(\frac{xx_1}{a^2}+\frac{yy_1}{b^2}\)

x(-1) + 8y(2) = 33

⇒ -x +16y = 33

⇒ x – 16y + 33 = 0

Equation of the normal is

16x + y + k = 0

It passes through P(-1, 2)

-16 + 2 + k = 0 ⇒ k =14

Equation of the normal is 16x + y + 14 = 0.

Question 2.

Find the equation of tangent and normal to the ellipse

x² + 2y² – 4x + 12y + 14 = 0 at (2, – 1).

Solution:

xx_{1} + 2yy_{1} – 2(x + x_{1}) + 6(y + y_{1}) + 14 = 0

⇒ 2x – 2y – 2(x + 2) + 6(y – 1) + 14 = 0

⇒ 4y + 4 = 0

y = – 1 required equation of tangent.

Slope of tangent is ‘0’

Equation of normal be

y + 1 = \(\frac{-1}{0}\) (x – 2)

x = 2 equation of normal.

Question 3.

Find the equation of the tangents to 9x² + 16y² = 144, which makes equal intercepts on the co-ordianate axis.

Solution:

Equation of the ellipse is 9x² + 16y² = 144

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

Equation of the tangent is

\(\frac{x}{a}\). cos θ + \(\frac{y}{b}\) sin θ = 1

Slope of the tangent = –\(\frac{b \cos \theta}{a \sin \theta}\) = -1

cot θ = \(\frac{a}{b}=\frac{4}{3}\)

cos θ = ± \(\frac{4}{5}\), sin 0 = ± \(\frac{3}{5}\)

Equation of the tangent is

\(\frac{x}{4}\)(±\(\frac{4}{5}\)) + \(\frac{y}{3}\)(±\(\frac{3}{5}\)) = 1

x ± y ± 5 = 0.

Question 4.

Find the co-ordinates for the points on the ellipse x² + 3y² = 37 at which the normal is parallel to the line 6x – 5y = 2.

Solution:

Equation of the ellipse is x² + 3y² = 37

⇒ \(\frac{x^2}{37}+\frac{y^2}{\left(\frac{37}{3}\right)}\) = 1

a² = 37, b² = \(\frac{37}{3}\)

Slope of the normal =

The normal is parallel to 6x – 5y = 2

Case i) The co-ordinates of P are

(a cos θ, b sin θ)

Case ii) The co-ordinates of P are (a cos θ, b sin θ)

Question 5.

Find the value of k if 4x+y + k = 0isa tangent to the ellipse x² + 3y² = 3.

Solution:

Equation of the ellipse is x² + 3y² = 3

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

a² = 3, b² = 1

Equation of the line is 4x + y + k = 0

y = -4x – k

m = -4, c = -k.

Condition for tangency is c² = a² m² + b²

(-k)² = 3(-4)² + 1

k²= 48 + 1 = 49

k = ±7.

Question 6.

Find the condition for the line x cos α + y sin α = p to be a tangent to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.

Solution:

Equation of the ellipse is

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) ……….. (1)

Equation of the line is x cos α + y sin α = p

y sin α = – x cos α + p cos α p

Condition for tangency is c² = a² m² + b²

II.

Question 1.

Find the equations of tangent and normal to the ellipse 2x² + 3y² = 11 at the point whose ordinate is 1.

Solution:

Equation of the ellipse is 2x² + 3y² = 11

Given y = 1

2×2 + 3 = 11 ⇒ 2x² = 8

x² = 4

x = ±2

Points on the ellipse are P (2, 1) and

Q(-2, 1)

Case i)P (2, 1)

Equation of the tangent is 2x.2 + 3y.1 = 11

4x + 3y = 11

The normal is perpendicular to the tangent Equation of the normal at P can be taken as 3x – 4y = k.

The normal passes through P (2, 1)

6 – 4 = k ⇒ k = 2

Equation of the normal at P is 3x – 4y = 2.

Case ii) Q (-2, 1)

Equation of the tangent at Q is

2x(-2) + 3y.1 = 11

– 4x + 3y = 11

4x – 3y + 11 = 0

Equation of the normal can be taken as

3x + 4y = k

The normal passes through Q (-2, 1)

-6 + 4 = k ⇒ k = -2

Equation of the normal at Q is 3x + 4y = -2

or 3x + 4y + 2 = 0.

Question 2.

Find the equations to the tangents to the ellipse, x² + 2y² = 3 drawn from the point (1, 2) and also find the angle between these tangents.

Solution:

Equations of the ellipse is x² + 2y² = 3

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

a² = 3, b² = \(\frac{3}{2}\)

Suppose m is the slope of the tangent. It passes through P (1, 2)

Equation of the tangent is

y – 2 = m(x – 1)

= mx – m

y = mx + (2 – m)

Condition for tangency is c² = a²m² + b²

(2 – m)² = 3(m²) + \(\frac{3}{2}\)

4 + m² – 4m = 3m² + \(\frac{3}{2}\)

2m² + 4m – \(\frac{5}{2}\)

4m² + 8m – 5 = 0

(2m – 1) (2m + 5) = 0

m = \(\frac{1}{2}\) or –\(\frac{5}{2}\)

Case i) m = \(\frac{1}{2}\)

Equation of the tangent is y = \(\frac{1}{2}\) x + 2 – \(\frac{1}{2}\)

= \(\frac{x}{2}+\frac{3}{2}\)

2y = x + 3

x- 2y + 3 = 0

Case ii) m = –\(\frac{5}{2}\)

Equation of the tangent is

y = – \(\frac{5}{2}\)x + (2 + \(\frac{5}{2}\))

= –\(\frac{5}{2}\)x + \(\frac{9}{2}\)

2y = – 5x + 9

or 5x + 2y – 9 = 0

Angle between the tangents is given by

Question 3.

Fincbthe equation of tangents to the ellipse 2x² + y² = 8 which are

i) Parallel to x – 2y – 4 = 0

Solution:

Slope will be : \(\frac{1}{2}\)

Equation of tangent y = mx ± \(\sqrt{a^2m^2+b^2}\)

2y – x + 6 = 0 required equation of tangents,

x – 2y ± 6 = 0.

ii) Perpendicular to x + y + 2 = 0

Solution:

Slope of tangent be ‘1’ as it is perpendicular to above line

y = mx ± \(\sqrt{a^2m^2+b^2}\)

y = x ± \(\sqrt{4+8}\)

y = x ± 2√3

⇒ x – y ± 2√3 = 0.

iii) Which makes an angle \(\frac{\pi}{4}\) with x-axis.

Solution:

Equation of tangent y = x ± 2√3.

⇒ x – y ± 2√3 = 0.

Question 4.

A circle of radius 4, is concentric with the ellipse 3x² + 13y² = 78. Prove that a common tangent is inclined to the major axis at an angle \(\frac{\pi}{4}\).

Solution:

Equation of the ellipse is 3x² + 13y² = 78

\(\frac{x^2}{26}+\frac{y^2}{6}\) = 1 ……… (1)

Centre of the ellipse is (0, 0)

Equation of the circle is x² + y² = 16 ………. (2)

Equation of tangent at P(θ) to the circle is

x cos θ + y sin θ = 16 ……. (3)

(3) is a tangent to the ellipse.

16 = 26 cos² θ + 6 sin² θ

= 26 cos² θ + 6(1 – cos² θ)

= 26 cos² θ + 6-6 cos² θ

20 cos² θ = 10

cos² θ = \(\frac{10}{20}=\frac{1}{2}\)

cos θ = ± \(\frac{1}{\sqrt{2}}\)

θ = \(\frac{\pi}{4}\)

III.

Question 1.

Show that the foot of the perpendicular drawn from the centre on any tangent to the ellipse lies on the curve

(x² + y²)² = a²x² + b²y².

Solution:

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

Equation of the tangent at P (θ) is

\(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1

CN is the perpendicular from C on the tangent slope of CN = \(\frac{y_1}{x_1}\)

∴ Slope of PT × slope of CN = -1

Locus of N (xv y,) is (x² + y²)² = a²x² + b²y²

Question 2.

Show that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle.

Solution:

Hint : Equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)

Equation of the tangent to the ellipse is

y = mx ± \(\sqrt{a^2m^2+b^2}\)

⇒ y – mx = ± \(\sqrt{a^2m^2+b^2}\) ………… (1)

Equation to the perpendicular from either focus (+ae, 0) on this tangent is

y = – \(\frac{1}{m}\) (x ± ae)

my = -(x ± ae)

my + x = ± ae ……….. (2)

Squaring and adding (1) and (2)

(y – mx)² + (my + x)² = a²m² + b² + a²e²

y² + m²x² – 2mxy + m²y² + x² + 2mxy = a²m² + a² – a²e² + a²e²

(x² + y²) (1 + m²) = a² (1 + m²)

⇒ x² + y² = a²

The locus is the auxiliary circle concentric with the ellipse.

Question 3.

The tangent and normal to the ellipse x² + 4y² = 4 at a point P(θ) on it meets the major axis in Q and R respectively. If 0 < θ < \(\frac{x}{2}\) and QR = 2, then show that θ = cos^{–}\(\frac{2}{3}\)

Solution:

Equation of the ellipse is x² + 4y² = 4

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

Equation of the tangent at P (θ) is

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

Equation of x-axis is y = 0

\(\frac{x}{2}\) cos θ = 1 ⇒ x = \(\frac{2}{\cos\theta}\)

Co-ordinates of Q are (\(\frac{2}{\cos\theta}\), θ)

Equation of the normal at

P(θ) is \(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}\) = a² – b²

\(\frac{ax}{\cos\theta}-\frac{by}{\sin\theta}\) = 3

Substituting y = 0 we get \(\frac{2x}{\cos\theta}\) = 3

x = \(\frac{3}{2}\). cos θ

Co-ordinates of R are (\(\frac{3}{2}\). cos θ, o)

Given QR = 2

\(\frac{-3 \cos ^2 \theta+4}{2 \cos \theta}\) = 2

-3 cos² θ+ 4 = 4 cos θ

3 cos² θ + 4 cos θ – 4 = 0

(3 cos θ – 2) ( cos θ + 2) = 0

3 cos θ – 2 = 0 or cos θ + 2 = 0

cos θ = \(\frac{2}{3}\) or cos θ = – 2

cos θ always lies between -1 and 1

∴ cos θ = \(\frac{2}{3}\)

i.e., θ = cos^{-1}(\(\frac{2}{3}\))