Class 11 RD Sharma Solutions - Chapter 26 Ellipse - Exercise 26.1 | Set 2

Chapter 26 of RD Sharma's Class 11 Mathematics textbook focuses on the ellipse a fundamental shape in the conic sections. This chapter aims to provide students with a comprehensive understanding of the properties, equations, and applications of ellipses. Exercise 26.1 Set 2 includes the problems designed to reinforce the concepts and skills related to the ellipses helping students to master the topic effectively.

Ellipse

An ellipse is a curve where the sum of the distances from any point on the curve to two fixed points called foci is constant. It can be described by the standard equation where a and b are the semi-major and semi-minor axes respectively. The Ellipses appear in various real-world applications including planetary orbits and optics. Understanding ellipses involves studying their geometric properties and relationships with the other conic sections.

Question 11. Find the equation of the ellipse whose foci are at (±3, 0) and which passes through (4, 1).

Solution:

Let the equation of the ellipse be  ....(i)

Given that the ellipse whose foci are at (±3, 0) and which passes through (4, 1)

So, 

ae = 3 

(ae)² = 9

y = 1 and  x = 4 

Substituting the values of x and y in the above equation, we have:

 ...(ii)

As we know that, b² = a²(1 - e²)

⇒ b² = a² - a²e²

⇒ b² = a² - 9 

or 

a² = b² + 9  ....(iii)

On solving eq(ii), we get

16b² + a²  = a²b²

Now put the value of a² from eq(iii), we get

16b² + b² + 9= (b² + 9)b²

b⁴ - 8b² - 9= 0

⇒ b = ±3

So, a = 3√2

Now put the value of a² and b² in eq(i), we get

Thus,is the required equation. 

Question 12. Find the equation of an ellipse whose eccentricity is 2/3, the latus rectum is 5 and the centre is the origin.

Solution:

Let the equation of the ellipse be  .....(i)

Given that 

eccentricity(e) = 2/3

latus rectum = 5 

So, 2b²/a = 5  .....(ii)

⇒ 2b² = 5a

⇒ 10a² = 45a

⇒ a = 9/2

On substituting the value of a in eq(ii), we have:

⇒ b² = 45/4

Now put the value of a² and b² in eq(i), we get

Thus,is the equation of the ellipse.

Question 13. Find the equation of the ellipse with its foci on the y-axis, eccentricity is 3/4, centre at the origin and passing through (6, 4).

Solution:

Let the equation of the plane be 

Given that 

eccentricity(e) = 3/4, 

As we know that  a² = b²(1 - e²)

⇒ 

⇒ 

⇒ 

Since it passes through (6,4), we have:

⇒ 

⇒ a² = 43 and b² = 688/7

Now put the value of a² and b² in eq(i), we get

Thus is the required equation.

Question 14. Find the equation of the ellipse whose axes lie along coordinate axes and which passes through (4, 3) and (-1, 4).

Solution:

Let the equation of the ellipse be:   ....(i)

It is given that the ellipse passes through (4, 3) and (-1, 4).

⇒ and 

Let and 

Then, 16p + 9r = 1 and p + 16b = 1

On solving these equations, we have:

and 

Now by substituting all the values in eq(i), we have:

Thus, is the required equation.

Question 15. Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point(-3, 1) and has an eccentricity equal to .

Solution:

Let the equation of the ellipse be:   ......(i)

It is given that the ellipse passes through the point (-3, 1).

So,

⇒ 

As we know that b² = a²(1 - e²)

Also the eccentricity(e) =  .

⇒ b² = a²(1 - 2/5)

⇒ b² = 3a²/5 

On substituting the values, we have:

⇒ a² = 32/2

⇒ b² = 32/5

Now put the value of a² and b² in eq(i), we get

Thus, 3x² + 5y² = 32is the required equation. 

Question 16. Find the equation of the ellipse, the distance between the foci is 8 units, and the distance between directrices is 18 units.

Solution:

Let the equation of the ellipse be:   ......(i)

Given that the distance between foci = 8 units

So, 2ae = 8  ...(i)

Distance between directrices = 18 units

So, 2a/e = 18  ...(i)

From eq(i) and (ii), we get

⇒ e = 8/2a

⇒ 4a² = 18(8)

⇒ a² = 36

⇒ a = 6

⇒ e = 2/3

Now, b² = a²(1 - e²)

⇒ b² = 36(1 - 4/9)

⇒ b² = 36(5/9)

⇒ b² = 20

Now put the values of a² and b² in eq(i), we get

Thus,is the required equation.

Question 17. Find the equation of the ellipse whose vertices are (0, ±10) and the eccentricity is 4/5.

Solution:

Let the equation of the ellipse be:   .....(i)

As we know that the vertices of the ellipse are on the y- axis, so the coordinates of the vertices are (0, ±10).

Thus, b = 10

Since, a² = b²(1 - e²)

eccentricity(e) = 4/5 (given) 

⇒ a² = 100(1 - (4/5)²) 

⇒ a² = 100(1 - 16/25) 

⇒ a² = 36

Now put the values of a² and b² in eq(i), we get

Thus,100x² + 36y²=3600is the required equation.

Question 18. A rod of length 12m moves with its ends touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with x-axis.

Solution:

Let us consider AB be the rod that make an angle θ with line OX and let P(x, y) be the point on it such that AP = 3 cm.

Then, PB = AB - AP = 12 - 3 = 9 cm

Thus, 

and, 

As we know that sin²θ  + cos²θ = 1, we have:

Thus is the locus of the point P. 

Question 19. Find the equation of the set of all points whose distances from (0, 4) are two-thirds of their distances from the line y = 9.

Solution:

Given that PQ = 2/3PL 

So,

⇒ 

⇒ 3²[x² + (y - 4)²] = 2²(y - 9)²

⇒ 9x² + 9y² - 72y + 144 = 4y² - 72y + 324

⇒ 9x² + 5y² = 180

Thusis the required equation.

Read More:

• Ellipse
• Ellipse Formula

Conclusion

Chapter 26 of RD Sharma's Class 11 Mathematics textbook delves into the properties and equations of the ellipses. Exercise 26.1 | Set 2 provides essential practice in understanding the standard forms of the ellipse deriving key parameters and applying the concepts to solve the various problems. Mastery of these exercises is crucial for the developing a strong foundation in conic sections which are vital in the advanced mathematical studies.