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KEAM 2017 Math Paper

Question : 22 of 120

Marks: +1, -0

The equation of the hyperbola with vertices (0,±15) and foci (0,± 20) is

Solution:

We have

Vertices and foci of hyperbola at

(0, \pm 15)

and

(0, \pm 20)

since, both foci and vertices lies on

Y

-axis, then equation of hyperbola will be

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

Now, vertices

= (0, \pm b)

\therefore \;\; b = 15

Again, foci

= (0, \pm b e)

\therefore \;\; b e = 20

\Rightarrow \;\; e = \frac{20}{15} = \frac{4}{3}

\Rightarrow \;\; \sqrt{1 + \frac{a^{2}}{b^{2}}} = \frac{4}{3}

\left[ \because e = \sqrt{1 + \frac{a^{2}}{b^{2}}} \right]

\Rightarrow \;\; 1 + \frac{a^{2}}{b^{2}} = \frac{16}{9}

\Rightarrow \;\; \frac{a^{2}}{b^{2}} = \frac{7}{9}

\Rightarrow \;\; a^{2} = \frac{7}{9} \times b^{2} = \frac{7}{9} \times 225

\Rightarrow \;\; a = \frac{\sqrt{7}}{3} \times 15 = 5 \sqrt{7}

\therefore

Equation of hyperbola is

\frac{y^{2}}{225} - \frac{x^{2}}{175} = 1

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