Question

The equations of the tangents to the hyperbola $x^2-9y^2=9$ that are drawn from $(3,2)$ and the area of the triangle that these tangents form with their chord of contact are:

• A. $y=\frac{5}{12}x+\frac{3}{4};y-2=0;8$sq.units
• B. $y=\frac{7}{12}x+\frac{5}{6};y-2=0;6$ sq.units
• C. $y=\frac{7}{12}x+\frac{5}{6};x-3=0;8$ sq.units
• D. $y=\frac{5}{12}x+\frac{3}{4};x-3=0;8$ sq. units

Answer 1

The correct option is D $y=\frac{5}{12}x+\frac{3}{4};x-3=0;8$ sq. units

Consider any line with equation $y=mx+c$. This line meets the hyperbola $x^2-{9x}^2=9$ at point whose $x$ coordinate are given by $x^2-9{(mx+c)}^2=9$
$\Rightarrow ({9m}^2-1)x^2+18mcx+9(1+c^2)=0$ ...(1)
If the line touches the curve, the roots of this equation are equal. In this case
${\left(18mc\right)}^2=36(9m^2-1)(1+c^2)\Rightarrow {9m}^2=1+c^2$ ...(2)
But the point $(3,2)$ is lies on line $y=mx+c\Rightarrow 2=3m+c$ ...(3)
From (2) and (3)
${9m}^2=1+{(2-3m)}^2\Rightarrow m=\frac{5}{12}$ or $m$ is infinity
If $m=\frac{5}{12}$, then $c=2-3m=\frac{3}{4}$
So the equation of one of the the tangent is $12y=5x+9$
If $m\to \infty$, the tangent is parallel to the $y$ axis, and as it passes through the point $(3,2)$ its equation is $x=3$
Now when the roots of equation (1) are equal
$x=\frac{18mc}{2({9m}^2-1)}$
on this put $m=\frac{5}{12},x=-5$ , the tangent $12y=5x+9$ touches the hyperbola at the point $(-5,\frac{-4}{5})$.
By observation , the other tangent touches the hyperbola at the vertex $(3,0)$
The triangle formed by the two angent $PA$ and $PB$ and chord of contact$AB$ has vertices $(3,2),(-5,\frac{-4}{3}),(3,0)$
The area $△$ of triangle $APB$ is given by
$2△==\left|\begin{array}{ccc}1& 1& 1\\ 3& -5& 3\\ 2& \frac{-4}{3}& 0\end{array}\right|\Rightarrow △=8$