OP Malhotra Class 11 Maths Solutions Chapter 24 Ellipse Chapter Test

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S Chand Class 11 ICSE Maths Solutions Chapter 24 Ellipse Chapter Test

Question 1.
Find the eccentricity of the ellipse
\(\frac{(x-3)^2}{8}\) + \(\frac{(y-4)^2}{6}\) = 1
Solution:
Eqn. of given ellipse be
\(\frac{(x-3)^2}{8}\) + \(\frac{(y-4)^2}{6}\) = 1 …(1)
which is an eqn. of horizontal ellipse Comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
We have a² = 8 and b² = 6
We know that b² = a² (1 – e²)
⇒ 6 = 8 (1 – e²) ⇒ e² = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
⇒ e = \(\frac{1}{2}\) (∵ e > 0)
Thus, required eccentricity of given ellipse be \(\frac{1}{2}\)

Question 2.
The distance between the foci of the ellipse 5x² + 9y² = 45 is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
Given eqn. be ellipse can be written as ;
\(\frac{x^2}{9}\) + \(\frac{y^2}{5}\) = 1 …(1)
which is a horizontal ellipse and
On comparing with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1,
where a > b > 0
We have, a² = 9 ; b² = 5
Also, b² = a²(1 – e²)
⇒ \(\frac{5}{9}\) = 1 – e²
⇒ e² = \(\frac{2}{3}\) (∵ e > 0)
Thus required distance between foci = 2ae
= 2 × 3 × \(\frac{2}{3}\) = 4 ∴ Ans. (c)

Question 3.
Find the equation of an ellipse whose latus rectum is 8 and eccentricity is \(\frac{1}{3}\)
Solution:
Given length of latus rectum of an ellipse = \(\frac{2 b^2}{a}\) = 8 ⇒ b² = 4a …(1)
and e = \(\frac{1}{3}\)
We know that b² = a² (1 – e²)
⇒ b² = a² \(\left[1-\frac{1}{9}\right]\) = \(\frac{8}{9}\)a²
⇒ 4a = \(\frac{8}{9}\)a²
⇒ a = \(\frac{36}{8}\) = \(\frac{9}{2}\)
∴ from (1); b² = 4 × \(\frac{9}{2}\) = 18
Thus required eqn. of an ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1; where a > b > 0
i.e. \(\frac{4 x^2}{81}\) + \(\frac{y^2}{18}\) = 1
⇒ 8x² + 9y² = 162

Question 4.
Find the equation of the ellipse whose foci are at (- 2, 4) and (4, 4) and major and minor axis are 10 and 8 respectively. Also, find the eccentricity of the ellipse.
Solution:
Since the foci of required ellipse are F (- 2, 4) and F (4, 4)
∴ distance between foci = 2ae = | FF’ |
= \(\sqrt{(4+2)^2+(4-4)^2}\) = 6
⇒ ae = 3 …(1)
Also length of major axis = 2a= 10 ⇒ a = 5
∴ from (1); e = \(\frac{3}{5}\) which gives the required eccentricity of an ellipse.
Since the y-coordinates of both points F and F’ are equal.
∴ major axis of the required ellipse is parallel to x-axis also the mid point of line segment FF’ gives the centre of required ellipse. Thus centre of ellipse be

Hence, the eqn. of ellipse having centre (1, 4) and major axis is parallel to x-axis is given by
\(\frac{(x-1)^2}{25}\) +\(\frac{(y-4)^2}{16}\) = 1
⇒ 16 (x – 1)² + 25 (y – 4)² = 400
⇒ 16x² + 25y² – 32x – 200y + 16 = 0

Question 5.
Find the equation of the ellipse whose 1 eccentricity is \(\frac { 1 }{ 2 }\) and whose foci are at the points (± 2, 0).
Solution:
Given foci of the ellipse be, (± 2, 0) and lies on x-axis. Thus x-axis be the major axis of an ellipse.
Let us take the eqn. of ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
Here ae = 2 and e = \(\frac { 1 }{ 2 }\)
⇒ a = 4
⇒ a² = 16
We know that b² = a²(1 -e²)
⇒ b² = 16 \(\left(1-\frac{1}{4}\right)\) = 16 × \(\frac { 3 }{ 4 }\) = 12
Thus eqn. (1) reduces to ; \(\frac{x^2}{16}\) + \(\frac{y^2}{12}\) = 1
which is the required eqn. of an ellipse.

Question 6.
Find the equation of the ellipse whose centre is the origin, major axis \(\frac { 9 }{ 2 }\) and eccentricity \(\frac{1}{\sqrt{3}}\) where the major axis is the horizontal axis.
Solution:
Since the major axis is the horizontal axis and let the eqn. of ellipse can be taken as
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
given 2a = \(\frac { 9 }{ 2 }\) ⇒ a = \(\frac { 9 }{ 4 }\) and e = \(\frac{1}{\sqrt{3}}\)
We know that b² = a² (1 – e²)
⇒ b² = \(\left(\frac{9}{4}\right)^2\) \(\left[1-\frac{1}{3}\right]\) = \(\frac{81}{16}\) × \(\frac{2}{3}\) = \(\frac{27}{8}\)
Thus eqn. (1) reduces to ;
\(\frac{16 x^2}{81}\) + \(\frac{8 y^2}{27}\) = 1
⇒ 16x² + 24y² = 81
which is the required eqn. of an ellipse.

Question 7.
Find the equation of the ellipse whose minor axis is 4 and which has a distance of 6 units between foci.
Solution:
Let us consider the eqn. of ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1; where a > b > 0 …(1)
given 2b = 4 ⇒ b = 2
Also distance between foci = 6
⇒ 2ae = 6 ⇒ ae = 3 …(2)
We know that b² = a² (1 – e²)
⇒ 4 = a² – 9 [using eqn. (2)]
⇒ a² = 13
Thus eqn. (1) reduces to; \(\frac{x^2}{13}\) + \(\frac{y^2}{4}\) = 1
which is the required eqn. of an ellipse.

Question 8.
Find the equation of the ellipse with its centre at (4, – 1), focus at (1, – 1) and given that i passes through (8, 0).
Solution:
Let S’ (α, β) be the other foci of the required ellipse. Thus C (4, – 1) be the mid-point of SS’.
∴ \(\frac{\alpha+1}{2}\) = 4 ⇒ α = 7
and \(\frac{\beta-1}{2}\) = – 1 ⇒ β = – 1
Thus the coordinates of other foci are S’ (7,-1).
Since the ordinate of points S and S’ are equal. Thus major axis is parallel to x-axis and it is a horizontal ellipse.
Let the eqn. of ellipse be
\(\frac{(x-4)^2}{a^2}\) + \(\frac{(y+1)^2}{b^2}\) = 1 …(1)
where a > b > 0
Distance between foci = 2ae = 6 ⇒ ae = 3
We know that b² – a² (1 – e²) = a² – 9
Thus eqn. (1) reduces to ;
\(\frac{(x-4)^2}{a^2}\) + \(\frac{(y+1)^2}{a^2-9}\) = 1 …(2)
eqn. (2) passes through the point (8, 0),
We get \(\frac{(8-4)^2}{a^2}\) + \(\frac{(y+1)^2}{a^2-9}\) = 1
⇒ \(\frac{16}{a^2}\) + \(\frac{1}{a^2-9}\) = 1
⇒ 16 (a² – 9) + a² = a² (a² – 9)
⇒ 17a² – 144 = a⁴ – 9a²
⇒ a⁴ – 26a² + 144 = 0
⇒ a⁴ – 18a² – 8a² + 144 = 0
⇒ a² (a² – 18) – 8 (a² – 18) = 0
⇒ (a² – 18) (a² – 8) = 0
⇒ a² = 18, 8
When a² = 8
∴ b² = a² – 9 = 8 – 9 = – 1
which is false
Thus a² = 18
∴ eqn. (2) reduces, to ;

Question 9.
Find the coordinates of the vertices and the foci and the length of the latus rectum of the ellipse 9x² + 25y² = 225.
Solution:
Given eqn. of ellipse can be written as
\(\frac{x^2}{25}\) + \(\frac{y^2}{9}\) = 1 …(7)
which is a horizontal ellipse.
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1,
where a > b > 0.
we have, a² = 25 and b² = 9
i.e. a = 5 and 6 = 3 (∵ a, b > 0)
Thus the coordinates of vertices are (± ae, 0) i.e. (± 5, 0).
We know that, b² = a² (1 – e²)
⇒ 9 = 25 (1 – e²) ⇒ 1 – e² = \(\frac{9}{25}\)
⇒ e² = \(\frac{16}{25}\) ⇒ e² = \(\frac{4}{5}\) (∵ e > 0)
Coordinates of foci are (± ae, 0)

and length of latus rectum = \(\frac{2 b^2}{a}\) = 2 × \(\frac{9}{5}\) = \(\frac{18}{5}\) units

Question 10.
Find the eccentricity and the equations of the directrices of the ellipse 7x² + 16y² = 112.
Solution:
Given eqn. of ellipse can be written as ;
\(\frac{x^2}{16}\) + \(\frac{y^2}{7}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
We have, a² = 16 b² = 7
We know that, b² = a² (1 – e²)
⇒ \(\frac{7}{16}\) = 1 – e² ⇒ e² = 1 – \(\frac{7}{16}\) = \(\frac{9}{16}\)
⇒ e = \(\frac{3}{4}\)
Thus required eqns. of directrices are given
by x = ± \(\frac{a}{e}\) = ± \(\frac{4 \times 4}{3}\) ⇒ 3x ± 16 = 0