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

  • Last Updated : 30 Jun, 2021

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

\frac{16}{a^2}+\frac{1}{b^2}=1 …(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

\frac{x^2}{18}+\frac{y^2}{9}=1



Thus, \frac{x^2}{18}+\frac{y^2}{9}=1      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 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 …..(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

\frac{4x^2}{81}+\frac{4y^2}{45}=1

Thus, \frac{4x^2}{81}+\frac{4y^2}{45}=1    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 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

Given that 

eccentricity(e) = 3/4, 

As we know that  a2 = b2(1 – e2)



⇒ a^2=b^2(1-\frac{9}{16})

⇒ a^2=\frac{7}{16}b^2

⇒ \frac{x^2}{a^2}+\frac{7y^2}{16a^2}=1

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

⇒ \frac{36}{a^2}+\frac{112}{16a^2}=1

⇒ a2 = 43 and b2 = 688/7

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

\frac{x^2}{43}+\frac{7y^2}{688}=1

Thus \frac{x^2}{43}+\frac{7y^2}{688}=1    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: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  ….(i)

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

⇒ \frac{16}{a^2}+\frac{9}{b^2}=1    and \frac{1}{a^2}+\frac{16}{b^2}=1

Let p=\frac{1}{a^2}    and r=\frac{1}{b^2}

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

On solving these equations, we have:

p=\frac{1}{a^2}=\frac{7}{247}    and r=\frac{1}{b^2}=\frac{15}{247}

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

\frac{7x^2}{247}+\frac{15y^2}{247}=1

Thus\frac{7x^2}{247}+\frac{15y^2}{247}=1    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 \sqrt{\frac{2}{5}} .

Solution:

Let the equation of the ellipse be: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  ……(i)

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

So,

⇒ \frac{9}{a^2}+\frac{1}{b^2}=1

As we know that b2 = a2(1 – e2)

Also the eccentricity(e) =  \sqrt{\frac{2}{5}} .

⇒ b2 = a2(1 – 2/5)

⇒ b2 = 3a2/5 

On substituting the values, we have:

\frac{9}{a^2}+\frac{5}{3a^2}=1

⇒ a2 = 32/2

⇒ b2 = 32/5

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

\frac{x^2}{\frac{32}{2}}+\frac{y^2}{\frac{32}{5}}=1

Thus3x2 + 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: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  ……(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

\frac{x^2}{36}+\frac{y^2}{20}=1

Thus, \frac{x^2}{36}+\frac{y^2}{20}=1    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: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  …..(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

\frac{x^2}{36}+\frac{y^2}{100}=1

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, cosθ=\frac{PQ}{PB}=\frac{x}{9}

and, sinθ=\frac{PR}{PA}=\frac{y}{3}

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

(\frac{y}{3})^2+(\frac{x}{9})^2=1

Thus \frac{x^2}{81}+\frac{y^2}{9}=1    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,

⇒ \sqrt{(x-0)^2+(y-4)^2}=\frac{2}{3}(y-9)

⇒ 32[x2 + (y – 4)2] = 22(y – 9)2

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

⇒ 9x2 + 5y2 = 180

Thus \frac{x^2}{20}+\frac{y^2}{36}=1    is the required equation.

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