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

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 = 32 is 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=3600 is 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

Thus is the required equation.