Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16 x$. If one of these tangents touches the two curves at $Q$ and $R$, then $(Q R)^2$ is equal to

Let be a point on the -axis. Common tangents are drawn

Let $A$ be a point on the $x$ -axis. Common tangents are drawn from $A$ to the curves $x^{2}+y^{2}=8$ and $\mathrm{y}^{2}=16 \mathrm{x}$ . If one of these tangents touches the two curves at $\mathrm{Q}$ and $\mathrm{R}$ , then $(\mathrm{QR})^{2}$ is equal to
Option: 1
81

Option: 2
72

Option: 3
76

Option: 4
64

Answers (1)

$y^{2}=16 x \quad$
Tangent $\quad x^{2}+y^{2}=8$
$y=m x+\frac{4}{m} \quad$ Tangent
$y=m x \pm 2 \sqrt{2} \sqrt{1+m^{2}}$
$\frac{4}{m}= \pm 2 \sqrt{2} \sqrt{1+m^{2}}$
$\frac{16}{\mathrm{~m}^{2}}=8+8 \mathrm{~m}^{2}$
$8 m^{4}+8 m^{2}=16$
$m^{4}+\mathrm{m}^{2}=2$
$m^{2}=1,-2$
Let $m=1$
$\underline{m>1}$
$\therefore \mathrm{y}=\mathrm{x}+4$

Point of tangecy at parabola
$Q\left(\frac{4}{m^{2}}, \frac{8}{m}\right)$
$Q(4,8)$

Point of tangency at circle eq^hat tangent at $R\left(x_{1}, y_{1}\right)$ is $\mathrm{T}=0$
$\mathrm{xx}_{1}+\mathrm{yy} \mathrm{y}_{1}=8$

Comparison with

$x-y+4=0$
$\frac{x_{1}}{1}=\frac{y_{1}}{-1}=-\frac{8}{4}$
$\mathrm{x}_{1}=-2 \quad \mathrm{y}_{1}=2$
$\mathrm{R}(-2,2)$
Now $Q R^{2}=\sqrt{(4+2)^{2}+(8-2)^{2}}$
$\mathrm{QR}^{2}=72$