Question

Determine the equation of pair of tangents that can be drawn from the point $(5,2)$ to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$.

• A. $9x^2-20xy-45x+8y+125=0$
• B. $13x^2-20xy-45x+50y-125=0$
• C. $13x^2+20xy-45x-50y+125=0$
• D. $13x^2-20xy-45x+50y+125=0$

Answer 1

The correct option is D $13x^2-20xy-45x+50y+125=0$

Given: Equation of hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1,$ coordinates of point $P=(5,2)$

To Find: Equation of pair of tangents

Step - $1$: Check whether the point $P$ lies outside or inside or on the curve.

Step - $2$: Recall formula of pair of tangents

Step - $3$: Substitute the values and simplify

From the given hyperbola, $a=5,b=3$

Checking if the point $P(5,2)$ lies outside the hyperbola or inside or on it.

$S=\frac{x^2}{25}-\frac{y^2}{9}-1$

$S_1=\frac{25}{25}-\frac{4}{9}-1=-\frac{4}{9}<0$

So, point $P(3,2)$ lies outside the hyperbola.

Now, equation of pair of tangents is $S{S}_1=T^2\cdots \left(1\right)$

$T=\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}-1$

$\Rightarrow T=\frac{5x}{25}-\frac{2y}{9}-1$

$\Rightarrow T=\frac{x}{5}-\frac{2y}{9}-1$

Substituting in $\left(1\right)$,
$(\frac{x^2}{25}-\frac{y^2}{9}-1)(-\frac{4}{9})=(\frac{x}{5}-\frac{2y}{9}-1{)}^2$

$[\because (a-b-c{)}^2=a^2+b^2+c^2-2ab+2bc-2ac]$

$(-\frac{4x^2}{225}+\frac{4y^2}{81}+\frac{4}{9})=\frac{x^2}{25}+\frac{4y^2}{81}+1-\frac{4xy}{45}+\frac{2y}{9}-\frac{x}{5}$

$\Rightarrow \frac{13x^2}{225}-\frac{4xy}{45}-\frac{x}{5}+\frac{2y}{9}+\frac{5}{9}=0$

Multiplying both sides by $225$, that is $\text{LCM}$ of $225,45,5,9$ we get,

$13x^2-20xy-45x+50y+125=0$