The area of the triangle formed by the tangen | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) Equation of chord of contact $AB $ is

$xh + yk = {a^2}$.....$(i)$

$OM = $length of perpendicular from $O(0, 0)$ on line $(i)$

$ = \frac{{

Vedclass · Accepted Answer

$a{\rm{ }}\frac{{{{({h^2} + {k^2} - {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$

Vedclass · Answer

(a) Equation of chord of contact $AB $ is

$xh + yk = {a^2}$.....$(i)$

$OM = $length of perpendicular from $O(0, 0)$ on line $(i)$

$ = \frac{{{a^2}}}{{\sqrt {{h^2} + {k^2}} }}$

$	herefore $ $AB = 2AM = 2\sqrt {O{A^2} - O{M^2}}  $

$= \frac{{2a\sqrt {{h^2} + {k^2} - {a^2}} }}{{\sqrt {{h^2} + {k^2}} }}$

Also $PM = $length of perpendicular from $P(h,\;k)$to the line $(i)$ is $\frac{{{h^2} + {k^2} - {a^2}}}{{\sqrt {{h^2} + {k^2}} }}$

Therefore, the required area of triangle $PAB$

$ = \frac{1}{2}.\;AB\;.\;PM $

$= \frac{{a{{({h^2} + {k^2} - {a^2})}^{3/2}}}}{{{h^2} + {k^2}}}$.

The area of the triangle formed by the tangents from the points $(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ and the line joining their points of contact is

Similar Questions

The normal at the point $(3, 4)$ on a circle cuts the circle at the point $(-1, -2)$. Then the equation of the circle is

The angle between the pair of tangents from the point $(1, 1/2)$ to the circle $x^2 + y^2 + 4x + 2y -4=0$ is-

If the tangent at $\left( {1,7} \right)$ to the curve ${x^2} = y - 6$ touches the circle ${x^2} + {y^2} + 16x + 12y + c = 0$ then the value of $c$ is:

Suppose two perpendicular tangents can be drawn from the origin to the circle $x^2+y^2-6 x-2 p y+17=0$, for some real $p$. Then, $|p|$ is equal to

The equation of the normal at the point $(4,-1)$ of the circle $x^2+y^2-40 x+10 y=153$ is