Question

The equations of the tangents to the hyperbola $4x^2-9y^2=1$ which are parallel to the line $4y=5x+7$ are

• A. $24y=30x\pm \sqrt{161}$
• B. $30x+24y=\pm \sqrt{161}$
• C. $30y=24\pm \sqrt{161}$
• D. $x+y=\sqrt{161}$

Answer 1

Answer: (A)

$24y=30x\pm \sqrt{161}$

Equation of hyperbola is $4x^2-9y^2=1$

or $\frac{x^2}{\frac{1}{4}}-\frac{y^2}{\frac{1}{9}}=1$ ...........$\left(1\right)$ which is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

where $a^2=\frac{1}{4}$ and $b^2=\frac{1}{9}$

Equation of the line parallel to the line $4y=5x+7$ or $5x-4y+7=0$ is $5x-4y+\lambda =0$

or $y=\frac{5}{4}x+\frac{\lambda }{4}$ ...........$\left(2\right)$

Slope of this line$=\frac{5}{4}=m$(let) and $c=\frac{\lambda }{4}$

If line$\left(2\right)$ touches the hyperbola $\left(1\right)$ then

$c^2=a^2{m}^2-b^2$

$\Rightarrow {\left(\frac{\lambda }{4}\right)}^2={\left(\frac{1}{2}\right)}^2{\left(\frac{5}{4}\right)}^2-{\left(\frac{1}{3}\right)}^2$

$\Rightarrow {\left(\frac{\lambda }{4}\right)}^2=\frac{1}{4}\times \frac{25}{16}-\frac{1}{9}=\frac{161}{64\times 9}$

$\Rightarrow {\lambda }^2=\frac{161}{36}\Rightarrow \lambda =\pm \frac{\sqrt{161}}{6}$

Substitute the values of $\lambda$ in equation$\left(2\right)$ the required equation of tangents are

$5x-4y\pm \frac{\sqrt{161}}{6}=0$

or $30x-24y\pm \sqrt{161}=0$