Question

A hyperbola has $y$-axis and $x$-axis as its conjugate axis and transverse axis respectively. If one of the points of intersection of $x$-axis with the hyperbola is $(4,0)$ and equation of one of the tangents is $x-y=\sqrt{7},$ then the equation of the hyperbola is

• A. $\frac{x^2}{16}-\frac{y^2}{9}=1$
• B. $\frac{x^2}{9}-\frac{y^2}{16}=1$
• C. $\frac{x^2}{25}-\frac{y^2}{9}=1$
• D. $\frac{x^2}{9}-\frac{y^2}{25}=1$

Answer 1

The correct option is A $\frac{x^2}{16}-\frac{y^2}{9}=1$

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Transverse axis is $x$-axis and conjugate axis is $y$-axis, so
$a\ge b$
Putting $(4,0)$
$a^2=16$
Equation of the hyperbola,
$\frac{x^2}{16}-\frac{y^2}{b^2}=1$

Now, equation of tangent to hyperbola is,
$y=mx\pm \sqrt{a^2{m}^2-b^2}$
Given equation of tangent is
$x-y=\sqrt{7}$
Comparing both the equation,
$\Rightarrow m=1$
Therefore, $x-\sqrt{7}=x\pm \sqrt{a^2\left(1\right)-b^2}$
$\Rightarrow -\sqrt{7}=\pm \sqrt{16-b^2}$
$\Rightarrow 7=16-b^2$
$\Rightarrow b^2=9$

Hence the equation of the hyperbola is
$\frac{x^2}{16}-\frac{y^2}{9}=1$