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JEE Advanced · Mathematics · 15. Hyperbola

Tangents are drawn to the hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\), parallel to the straight line \(2 x-y=1\). The points of contact of the tangents on the hyperbola are

  1. A \(\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
  2. B \(\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)\)
  3. C \((3 \sqrt{3},-2 \sqrt{2})\)
  4. D \((-3 \sqrt{3}, 2 \sqrt{2})\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)\)

Step-by-step Solution

Detailed explanation

If slope of tangents to hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is \(m\),

then equations of tangent to the hyperbola is

\(y=m x \pm \sqrt{a^{2} m^{2}-b^{2}} \quad\) with the points of contact

\(\quad\left(\quad \pm a^{2} m\right.\)

\(\left.\frac{\pm \sqrt{a^{2} m^{2}-b^{2}}}{\sqrt{a^{2} m^{2}-b^{2}}}\right)\)

\(\therefore\) Tangent to hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\) is parallel to \(2 x-y=1\),

\(\therefore\) Slope of tangent \(=2\)

\(\therefore\) Points of contact are \(\left(\frac{\pm 9 \times 2}{\sqrt{9 \times 4-4}}, \frac{\pm 4}{\sqrt{9 \times 4-4}}\right)\)

i.e. \(\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) and \(\left(\frac{-9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)\)
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