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Exercise 10.4 · Q3

Q.Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola 9y2−4x2=369y^2 - 4x^2 = 36.

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Rewrite the hyperbola in standard form y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 to identify its orientation (vertical axis), then use the relationships c2=a2+b2c^2 = a^2 + b^2 and e=cae = \frac{c}{a} to find all elements. Vertices: (0,±2)(0, \pm 2); Foci: (0,±13)(0, \pm \sqrt{13}); Eccentricity: 132\frac{\sqrt{13}}{2}; Latus rectum: 99.

The equation 9y2−4x2=369y^2 - 4x^2 = 36 describes a hyperbola, but not in its most useful form. To extract geometric information—where the vertices and foci sit, how "stretched" the hyperbola is—we need the standard form. For a hyperbola, this means getting 11 on the right-hand side and identifying which variable is positive (that determines the axis).

Dividing the entire equation by 3636:

9y236−4x236=1\frac{9y^2}{36} - \frac{4x^2}{36} = 1

y24−x29=1\frac{y^2}{4} - \frac{x^2}{9} = 1

This is now in standard form. Because the y2y^2 term is positive, the hyperbola opens vertically (along the yy-axis). The standard template for such a hyperbola centered at the origin is:

y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

where a2=4a^2 = 4 and b2=9b^2 = 9.

For a vertical hyperbola y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1:

  • Vertices: (0,±a)(0, \pm a)
  • Foci: (0,±c)(0, \pm c) where c2=a2+b2c^2 = a^2 + b^2
  • Eccentricity: e=cae = \frac{c}{a}
  • Latus rectum: 2b2a\frac{2b^2}{a}

Now we extract each piece systematically.

Finding the vertices

The vertices lie on the transverse axis at a distance aa from the center. Here a2=4a^2 = 4, so a=2a = 2.

Vertices: (0,2)(0, 2) and (0,−2)(0, -2).

Finding the foci

The foci lie further out along the same axis. The relationship for hyperbolas is c2=a2+b2c^2 = a^2 + b^2 (note the plus sign, unlike ellipses).

c2=4+9=13c^2 = 4 + 9 = 13

c=13c = \sqrt{13}

Foci: (0,13)(0, \sqrt{13}) and (0,−13)(0, -\sqrt{13}).

Watch out

Students often confuse the hyperbola relation c2=a2+b2c^2 = a^2 + b^2 with the ellipse relation c2=a2−b2c^2 = a^2 - b^2. For hyperbolas, the focal distance is larger than the semi-transverse axis, so we add.

Finding the eccentricity

Eccentricity measures how "open" the hyperbola is. For any hyperbola, e>1e > 1.

e=ca=132e = \frac{c}{a} = \frac{\sqrt{13}}{2}

This value is approximately 1.81.8, confirming the hyperbola is moderately open.

Finding the length of the latus rectum

The latus rectum is the chord through a focus, perpendicular to the transverse axis. Its length is given by:

Latus rectum=2b2a=2⋅92=9\text{Latus rectum} = \frac{2b^2}{a} = \frac{2 \cdot 9}{2} = 9

Tip

The latus rectum formula 2b2a\frac{2b^2}{a} works for both parabolas (where it simplifies to 4a4a) and hyperbolas. It's a measure of the "width" of the conic at the focus.

ElementValue
Vertices(0,±2)(0, \pm 2)
Foci(0,±13)(0, \pm \sqrt{13})
Eccentricity132\frac{\sqrt{13}}{2}
Latus rectum99
✓Final answer

The vertices are (0,±2)(0, \pm 2), the foci are (0,±13)(0, \pm \sqrt{13}), the eccentricity is 132\frac{\sqrt{13}}{2}, and the length of the latus rectum is 99.

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