Question

Focus of hyperbola is $(\pm 3,0)$ and equation of tangent is $2x+y-4=0$, find the equation of hyperbola is

• A. $4x^2–5y^2=20$
• B. $5x^2–4y^2=20$
• C. $4x^2–5y^2=1$
• D. $5x^2–4y^2=1$

Answer 1

The correct option is A

$4x^2–5y^2=20$

Explanation for correct option:

Step1. General equation hyperbola

General equation of hyperbola is given by

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)$,also $b^2=a^2(e^2-1)$

Step2. Finding value of $a\&b$

Given, Foci of the hyperbola is $(\pm 3,0)$ i.e. $ae=\pm 3$

$\begin{array}{rcl}{b}^2& =& a^2(e^2-1)\\ b^2& =& {\left(ae\right)}^2-a^2\\ b^2& =& 3^2-a^2\\ a^2+b^2& =& 9...\left(ii\right)\end{array}$

Equation of tangent on hyperbola is $2x+y-4=0$ touches equation $\left(i\right)$ if $a^2\times 2^2-b^2\times 1^2=4^2...\left(iii\right)$

Now, Equation of tangent on curve is given by $lx+my+n=0$touches equation $\left(i\right)$if $a^2{l}^2-b^2{m}^2=n^2$

Add equation $\left(ii\right)\&\left(iii\right)$

$\begin{array}{rcl}{a}^2+b^2& =& 9\\ 4a^2-b^2& =& 16\\ & & -------\\ 5a^2& =& 25\\ a^2& =& 5\&b^2=4\\ & & \end{array}$

Step3. finding equation hyperbola.

Put the value of $a^2\&b^2$ in equation $\left(i\right)$

Equation of parabola is

$\begin{array}{rcl}\frac{x^2}{5}-\frac{y^2}{4}& =& 1\\ 4x^2-5y^2& =& 20\end{array}$

Hence correct option is $\left(A\right)$