Tag: maths

Let $P$ be $(h, k)$ then by $SS _{1} = T^{2}$ the equation of pair of tangents drawn from $P$ to $x^{2} + y^{2} = 1$ is
$(h^{2} + k^{2} - 1)(x^{2} + y^{2} - 1) = (hx + ky - 1)^{2}$
Its intersection with the line $x = 1$ is given by
$y^{2}(h^{2} + k^{2} - 1) = (ky + h - 1)^{2}$
or $y^{2}(h^{2} - 1) - 2yk (h - 1) - (h - 1)^{2} = 0 .....(1)$
It is a quadratic in $y$ and we are given that length of intercept is $1 \therefore y _{1} - y _{2} = 2$
or $(y _{1} + y _{2})^{2} - 4y _{1}y _{2} = 1$
or $\left [\dfrac {2k(h - 1)}{h^{2} - 1}\right ]^{2} + 4\dfrac {(h - 1)^{2}}{h^{2} - 1} = 4$
or $\dfrac {4k^{2}}{(h + 1)^{2}} + \dfrac {4(h - 1)}{h + 1} = 4$
or $k^{2} = (h + 1)^{2} - (h^{2} - 1) = 2h + 2$
Hence the locus of $(h, k)$ is $y^{2} =2(x + 1)$ which represents a parabola.

Given function,

$f(x)=1+c\left( x+3 \right)$                     .....( 1 )               where $c\in R$

We know that,

General equation of circle of radius $a$ meets at the origin is

${{x}^{2}}+{{y}^{2}}={{a}^{2}}$               

But it is unit circle then $a=1$,

Then, equation of circle is

$ {{x}^{2}}+{{y}^{2}}={{1}^{2}} $

$ {{x}^{2}}+{{y}^{2}}=1 $

Let  $f\left( x \right)=y$

By equation $\left( 1 \right)$ and we get,

$y=1+c\left( x+3 \right)$

$y=1+cx+3$                          ......( 2 )

Using distance formula,   at the origin to a line

$ d=\left| \dfrac{0-c\times 0-1-3c}{\sqrt{{{1}^{2}}+{{c}^{2}}}} \right|=1 $

$ \left| \dfrac{-1-3c}{\sqrt{{{1}^{2}}+{{c}^{2}}}} \right|=1 $

Taking square both side and we get, 

$ {{\left( \dfrac{1+3c}{\sqrt{1+{{c}^{2}}}} \right)}^{2}}=1 $

$ {{\left( 1+3c \right)}^{2}}=1+{{c}^{2}} $

$ {{1}^{2}}+{{\left( 3c \right)}^{2}}+6c=1+{{c}^{2}} $

$ 1+9{{c}^{2}}+6c=1+{{c}^{2}} $

$ 9{{c}^{2}}+6c-{{c}^{2}}=0 $

$ 8{{c}^{2}}+6c=0 $

$ 2c\left( 4c+3 \right)=0 $

$ 2c=0,4c+3=0 $

$ c=0,4c=-3 $

$ c=0,c=-\dfrac{3}{4} $

 Hence, It is required solution.

Since tangent is perpendicular to the radius through the point of contact
so, $PQ \bot OQ$