If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one of its foci be at (–3, 0), then which one of the following points does not lie on this hyperbola?

Hyperbola PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q30 with Solution

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Correct answer: A

Find the equation of the hyperbola

The vertices are at $(-2,0)$ and $(2,0)$ , so the center is at $(0,0)$ and the transverse axis is along the $x$ -axis.

Hence the standard form is

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

Since the vertices are $(\pm 2,0)$ , we get $a=2 \implies a^2=4.$

One focus is at $(-3,0)$ , so $c=3 \implies c^2=9.$

For a hyperbola, $c^2=a^2+b^2.$ Thus, $9=4+b^2 \implies b^2=5.$

Therefore, the equation of the hyperbola is

\frac{x^2}{4}-\frac{y^2}{5}=1.

Check each option

We test whether each point satisfies

\frac{x^2}{4}-\frac{y^2}{5}=1.

Option A: $\left(6,5\sqrt{2}\right)$

\frac{6^2}{4}-\frac{(5\sqrt{2})^2}{5} =\frac{36}{4}-\frac{50}{5} =9-10=-1.

This is not equal to $1$ .

So, $\left(6,5\sqrt{2}\right)$ does not lie on the hyperbola.

Option B: $\left(2\sqrt{6},5\right)$

\frac{(2\sqrt{6})^2}{4}-\frac{5^2}{5} =\frac{24}{4}-\frac{25}{5} =6-5=1.

This lies on the hyperbola.

Option C: $\left(-6,2\sqrt{10}\right)$

\frac{(-6)^2}{4}-\frac{(2\sqrt{10})^2}{5} =\frac{36}{4}-\frac{40}{5} =9-8=1.

This lies on the hyperbola.

Option D: $\left(4,\sqrt{15}\right)$

\frac{4^2}{4}-\frac{(\sqrt{15})^2}{5} =\frac{16}{4}-\frac{15}{5} =4-3=1.

This lies on the hyperbola.

Conclusion

The point that does not lie on the hyperbola is:

\boxed{\left(6,5\sqrt{2}\right)}

which is Option A.

More from Hyperbola

View written solutionFree

Correct answer: A

Find the equation of the hyperbola

The vertices are at $(-2,0)$ and $(2,0)$ , so the center is at $(0,0)$ and the transverse axis is along the $x$ -axis.

Hence the standard form is

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

Since the vertices are $(\pm 2,0)$ , we get $a=2 \implies a^2=4.$

One focus is at $(-3,0)$ , so $c=3 \implies c^2=9.$

For a hyperbola, $c^2=a^2+b^2.$ Thus, $9=4+b^2 \implies b^2=5.$

Therefore, the equation of the hyperbola is

\frac{x^2}{4}-\frac{y^2}{5}=1.

Check each option

We test whether each point satisfies

\frac{x^2}{4}-\frac{y^2}{5}=1.

Option A: $\left(6,5\sqrt{2}\right)$

\frac{6^2}{4}-\frac{(5\sqrt{2})^2}{5} =\frac{36}{4}-\frac{50}{5} =9-10=-1.

This is not equal to $1$ .

So, $\left(6,5\sqrt{2}\right)$ does not lie on the hyperbola.

Option B: $\left(2\sqrt{6},5\right)$

\frac{(2\sqrt{6})^2}{4}-\frac{5^2}{5} =\frac{24}{4}-\frac{25}{5} =6-5=1.

This lies on the hyperbola.

Option C: $\left(-6,2\sqrt{10}\right)$

\frac{(-6)^2}{4}-\frac{(2\sqrt{10})^2}{5} =\frac{36}{4}-\frac{40}{5} =9-8=1.

This lies on the hyperbola.

Option D: $\left(4,\sqrt{15}\right)$

\frac{4^2}{4}-\frac{(\sqrt{15})^2}{5} =\frac{16}{4}-\frac{15}{5} =4-3=1.

This lies on the hyperbola.

Conclusion

The point that does not lie on the hyperbola is:

\boxed{\left(6,5\sqrt{2}\right)}

which is Option A.

Hyperbola question

Hyperbola question

Option A: $\left(6,5\sqrt{2}\right)$

Option B: $\left(2\sqrt{6},5\right)$

Option C: $\left(-6,2\sqrt{10}\right)$

Option D: $\left(4,\sqrt{15}\right)$

More from Hyperbola

Hyperbola question

Hyperbola question

Option A: $\left(6,5\sqrt{2}\right)$

Option B: $\left(2\sqrt{6},5\right)$

Option C: $\left(-6,2\sqrt{10}\right)$

Option D: $\left(4,\sqrt{15}\right)$

More from Hyperbola

Option A: (6,52)\left(6,5\sqrt{2}\right)(6,52​)

Option B: (26,5)\left(2\sqrt{6},5\right)(26​,5)

Option C: (−6,210)\left(-6,2\sqrt{10}\right)(−6,210​)

Option D: (4,15)\left(4,\sqrt{15}\right)(4,15​)

More from Hyperbola

Option A: (6,52)\left(6,5\sqrt{2}\right)(6,52​)

Option B: (26,5)\left(2\sqrt{6},5\right)(26​,5)

Option C: (−6,210)\left(-6,2\sqrt{10}\right)(−6,210​)

Option D: (4,15)\left(4,\sqrt{15}\right)(4,15​)

More from Hyperbola

Option A: $\left(6,5\sqrt{2}\right)$

Option B: $\left(2\sqrt{6},5\right)$

Option C: $\left(-6,2\sqrt{10}\right)$

Option D: $\left(4,\sqrt{15}\right)$

Option A: $\left(6,5\sqrt{2}\right)$

Option B: $\left(2\sqrt{6},5\right)$

Option C: $\left(-6,2\sqrt{10}\right)$

Option D: $\left(4,\sqrt{15}\right)$